PHYS 1100 Newton's Laws Solutions

Physics 1100: 2D Kinematics Solutions

In the diagrams below, a ball is on a flat horizontal surface. The initial velocity and constant external forces acting on the ball are indicated. Describe qualitatively how motion the motion of the ball will change.

First determine the direction of the net force on the ball. From Newton's Second Law, F = ma, so the direction of the net force and resulting acceleration are in the same direction. As time passes, the direction of the ball will tend to point in the direction of the acceleration.

Three forces F₁ = (25.0N, 42.5°), F₂= (15.5N, 215°), and F₃ = (20.5N, 155°) accelerate an 8.75 kg mass. What is the net force acting on the mass? What is the magnitude and direction of the mass's acceleration?

This involves a 2D vector addition, so it is appropriate to sketch the addition of the forces and then add the components:

F₁ = i[25cos(42.5°)] + j[25sin(42.5°)] = i[18.432] + j[16.890]

F₂ = i[-15.5cos(35.0°)] + j[-15.5sin(35.0°)] = i[-12.697] + j[-8.8904]

F₃ = i[-20.5sin(65.0°] + j[-20.5cos(65.0°] = i[-18.579] + j[8.664]

Adding the components of the vectors we get

F_net

=

i[18.432 + -12.697 + -18.579] + j[16.890 + -8.8904 + 8.664]

=

i[-12.844] + j[16.663]

The magnitude of the net force is given by F_net = [(F_x)² + (F_y)²]^½ = 21.039 N . The angle is determined by trigonometry to be θ = arctan(|F_y/F_x|) = 52.4°. So putting the answer in the same form as that we were given, the net force is F_net= (21.0 N, 127.6°).

Since a = F_net/m, a = ([21.0 N / 8.75 kg], 127.6°) = (2.40 m/s², 127.6°).

What would have to be the magnitude and direction of a fourth force F₄ in question 2 so that the acceleration of the mass would be zero?

We are asked to find F₄ such that F₁ + F₂ + F₃+ F₄ = F_net + F₄= 0. In other words, we find to find F₄=−F_net. Equal and opposite vectors have the same magnitude but are 180° apart, so F₄= (21.0 N, 127.6° + 180°) = (21.0 N, 307.6°).

A rifle bullet with a mass of 100 g leaves the rifle with a muzzle velocity of 500 m/s. If the barrel is 0.75 m long, calculate the average acceleration of the bullet. Calculate the average force that the exploding gunpowder exerts on the bullet. If the rifle has a mass of 7.50 kg, what is its average acceleration?

(a) To calculate the average acceleration, we make use of the kinematic data. To solve a kinematics problem, we list the given data and the unknown. We may reasonably assume that the bullet starts from rest.

v₀	= 0
v_f	= 500 m/s
Δx	= 0.75 m
a	= ?

Examining the variables, we see that we can use the equation 2aΔx = (v)² − (v₀)², to find the acceleration,

The plus sign indicates that acceleration is in the same direction as the motion of the bullet.

(b) According to Newton's Second Law,

F_bullet = m_bulleta_bullet = (0.100 kg)(1.667 × 10⁵ m/s²) = +1.667 × 10⁴ N ,

where the plus sign indicates that the force is in the same direction as the motion of the bullet.

(c) According to Newton's Third Law, the force on the rifle must be equal but opposite to the force acting on the bullet, F_rifle = −F_bullet. Hence we know that the force on the rifle is −1.667× 10⁴ N, where the plus sign indicates that the force is in the opposite direction to the motion of the bullet.

(d) According to Newton's Second Law, F_rifle = m_riflea_rifle. We thus find the accelartation of the bullet to be

a_rifle = F_rifle/m_rifle = (−1.667 × 10⁴ N) / (7.50 kg) = −2.22 × 10³ m/s² .

where the minus sign indicates that the acceleration of the rifle is in the opposite direction to the motion of the bullet.

You are thrown from a bicycle and skid in a straight line to a complete stop on a rough gravel path. A measurement of the bloody skidmark reveals that it is 3.50 m long. What average force did the gravel exert on your anatomy? Assume that your mass is 70.0 kg and that your initial speed was 30 km/h.

To find the force we would use Newton's Second Law, F = ma. We are given the mass but not the acceleration. However we are given kinematic data and we should be able to use this to find the acceleration.

v₀ = 0

v_f = 30 km/h (1000 m / 1 km) (1h / 3600 s)
= 8.333 m/s

Δx = 3.50 m

a = ?

Examining the variables, we see that we can use the equation 2aΔx = (v)² − (v₀)², to find the acceleration,

The minus sign indicates that acceleration is in the opposite direction to your motion.

According to Newton's Second Law,

F = ma = (70.0 kg)(−9.92 m/s²) = −694 N ,

where the minus sign indicates that the force is also in the opposite direction to your motion as you slide.

The 80.0-kg male partner of a figure skating duo pushes his 60.0-kg female partner with a force of 70.0N. Find the acceleration of both partners.

According to Newton's Third Law, the woman must exert 70.0 N on the male skater but in the opposite direction, in other words, −70.0 N, where the minus sign indicates the direction opposite to the direction that the man pushes the woman.

To find the acceleration of the male skater we use Newton's Second Law,

a_man = F_womanonman / m_man = −70 N / 80.0 kg = −0.875 m/s².

The minus sign indicates that the man moves backwards which is what one would expect.

Similarly the acceleration of the female skater is

a_woman = F_manonwoman / m_woman = +70 N / 60.0 kg = +1.17 m/s².

The plus sign indicates that the woman moves backwards (forward from the man's perspective) which is what one would expect.

An 800-kg artillery piece horizontally fires a 10.5-kg shell with a muzzle velocity of 450 m/s. The barrel is 3.75 m long. What is the acceleration of both?

To calculate the acceleration of the shell, we make use of the kinematic data. To solve a kinematics problem, we list the given data and the unknown. We may reasonably assume that the shell starts from rest.

v₀ = 0

v_f = 450 m/s

Δx = 3.75 m

a = ?

Examining the variables, we see that we can use the equation 2aΔx = (v)² − (v₀)², to find the acceleration,

The plus sign indicates that acceleration is in the same direction as the motion of the shell.

We don't have any kinematic data for the artillery piece, but we do know from Newton's Third Law that the force on the cannon must be equal but opposite to the force acting on the shell, F_cannon = −F_shell, and we can calculate F_shell.

According to Newton's Second Law,

F_shell = m_shella_shell = (10.5 kg)(2.70 × 10⁴ m/s²) = +2.84 10⁵ N ,

where the plus sign indicates that the force is in the same direction as the motion of the shell.

Since F_cannon = m_cannona_cannon, we rearrange to find

a_cannon = F_cannon/m_cannon = −F_shell/m_cannon = (−2.84 × 10⁵ N)/(800 kg) = −354 m/s² .

where the minus sign indicates that the acceleration of the cannon is in the opposite direction to the acceleration of the shell.

If m₁ = 25.0 kg, calculate the normal force on the bottom mass if the applied force, F, is (a) 100 N, (b) 245 N, and (c) 500 N.

Since this problem deals with forces that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for the block. The block has mass, so it has weight. The table top exerts a normal on the block. A given force is also acting on the block. The table top normal is nonzero only if the block remains on the table, so we will assume that the acceleration is zero.

In component form the forces and acceleration vectors are;

F = i0 + jF,

N_table = i0 – jN_table,

W = i0 – jmg, and

a = i0 + j0.

Since ΣF_y = ma_y, we find N_table + F - mg = 0. We don’t find anything interesting from ΣF_x = ma_x. So the equation for the normal force is

N_table = mg - F .

So the equation for the normal force is

N_table = mg − F .

Examining the three cases yields,

F N_table

(a) 100 N 145 N

(b) 245 N 0 N

(c) 500 N −255 N

Since normal force can never be zero, case (c) is saying that the block has been lifted off the table, so that N_table = 0, and the block is actually accelerating upwards.

What is the normal force on the object of mass m₁ shown in the diagram? What is the acceleration of the object? In the diagram F₁ = 25 N, F₂ = 15 N, and m₁ = 20 kg.

This problem deals with forces and acceleration, which suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). In the diagram we show the givens forces F₁ and F₂. As well, since the object has mass, it has weight. Since the object touches the table top, there is a normal force from the table top through the object. We assume that the object will accelerate to the right.

Next we break the forces and acceleration into components;

F₁ = iF₁cos(35°) – jF₁sin(35°),

F₂ = iF₂cos(43°) – jF₂sin(43°),

N = i0 – jN,

W = i0 – jmg, and

a = ia + j0.

We apply Newton’s Second Law separately to the i and j components. Thus ΣF_x = ma_x yields

F₁cos(35°) + F₂cos(43°) = m₁a

and ΣF_y = ma_y gives

N - m₁g - F₁sin(35°) + F₂sin(43°) = 0

Thus the normal force is:

N = m₁g + F₁sin(35°) − F₂sin(43°) = (20)(9.81) +25sin(35°)− 15sin(43°) = 200.3 N

The acceleration is:

a = [F₁cos(35°) + F₂cos(43°)]/m₁ = [25cos(35°) + 15cos(43°)]/20 = 1.572 m/s².

A physics instructor is pulling a string with a constant force of 1.00 Newtons. The string is attached to a toy on a frictionless surface. The string makes a 50º angle with the horizontal. The toy has a mass of 0.400 kg. Find the acceleration of the toy.

This problem deals with forces and acceleration, which suggests we should apply Newton’s Second Law. To apply Newton’s Second Law we draw a free-body diagram (FBD). In the diagram we show the given force F. As well, since the object has mass, it has weight. Since the object touches the tabletop, there is a normal force from the tabletop through the object. We assume that the object will accelerate to the right.

We break the forces and acceleration into components;

F = iFcos(50°) – jFsin(50°),

N = i0 – jN,

W = i0 – jmg, and

a = ia + j0.

We apply Newton’s Second Law separately to the i and j components. Thus ΣF_x = ma_x yields

Fcos(50°) = ma

and ΣF_y = ma_y gives

N - mg + Fsin(50°) = 0

Thus the normal force is:

N = mg − Fsin(50°) = (0.400)(9.81) − 1.00sin(50°) = 3.16 N

The acceleration is:

a = Fcos(50°)/m = 1.00cos(50°)/0.400 = 1.61 m/s² .

Since the acceleration is positive, our assumption about its direction was correct. The fact that the normal is positive, i.e. that the object has not lost contact with the tabletop, tells us that there is no acceleration in the y direction.

Two blocks are connected by a string. The blocks have masses of 1.50 kg and 1.20 kg. The heavier block is being pulled by a force of 8.00 N at angle of 37° to the floor. Assume the floor is frictionless. Find the tension in the string between the blocks.

This problem deals with forces and acceleration, which suggests we should apply Newton’s Second Law. To apply Newton’s Second Law we draw a free-body diagram (FBD) for each block. In the diagram for m₁ we show the given force F. As well, since the object has mass, it has weight. Since the object touches the tabletop, there is a normal force from the tabletop through the object. There is a string attached to m₁ so there is a tension from the heavier block to the lighter. In the diagram for m₂ we show the tension of the string acting on it which is the same as on m₁ but opposite in direction. It also has weight and is on a surface. Both objects should have the same acceleration since they are connected by a string. We will assume that they will accelerate to the right.

For each block, we break the forces and acceleration into components

Left Block

Right Block

F = iFcos(37°) + jFsin(37°)

T = iT + j0

T = –iT + j0

N₂ = i0 – jN₂

N₁ = i0 – jN₁

W₂ = i0 – jm₂g

W₁ = i0 – jm₁g

a₂ = ia + j0

a₁ = ia + j0

We apply Newton's Second Law separately to the i and j components for each block.

Object

i

j

ΣF_x = ma_x

ΣF_y = ma_y

Right Block

Fcos(37°) - T = m₁a

N₁ - m₁g + Fsin(37°) = 0

Left Block

T = m₂a

N₂ - m₂g = 0

Thus the normal forces are:

N₁ = m₁g − Fsin(50°) = (1.50)(9.81) − 8.00sin(37°) = 9.90 N ,

and

N₂ = m₂g = (1.20)(9.81) = 11.77 N .

The fact that both normals are positive indicates that the blocks do not lose contact with the tabletop and are therefore not accelerating in the y direction. This is consistent with our assumption about the acceleration being only in the x direction.

To find the tension T we need to eliminate the acceleration a from the pair of x equations. The second gives

a = T / m₂.

We substitute this into the first equation

Fcos(37°) − T = m₁ [T / m₂] .

We put terms with T on one side and everything else on the other side

Fcos(37°) = m₁ [T / m₂] + T .

We collect T and find

Fcos(37°) = T [m₁/m₂ + 1] .

So we have

T = Fcos(37°)/[m₁/m₂ + 1] = 8.00cos(37°)/[1.50/1.20 + 1] = 2.84 N .

So the tension in the string between the blocks is 2.84 N.

The acceleration of the blocks is

a = T/m₂ = 2.84 / 1.20 = 2.37 m/s² .

This is positive so our assumption about the direction of the acceleration was correct.

The acceleration is:

a = Fcos(50°)/m = 1.00cos(50°)/0.400 = 1.61 m/s² .

Two blocks are connected by strings to the ceiling as shown in the diagram below. The left block has mass m₂ = 3.35 kg. The mass of the right block m₁ is unknown. The strings make angles θ = 60º and φ = 45º with the ceiling. Find the tension in the level string connecting the blocks. Find m₁.

This problem deals with a force, tension, which suggests we should apply Newton’s Second Law. To apply Newton’s Second Law we draw a free-body diagram (FBD) for each block. In the diagram for m₁ we show two tension forces, one for each string. As well, since the object has mass, it has weight. In the diagram for m₂ we again show two tension forces and its weight. We would not expect the blocks to move so we will assume that there is no acceleration.

In the free-body diagrams, we also use a little geometry to find more useful angles to work with.

For each block, we break the forces and acceleration into components

Left Block	Right Block
T₂ = –iT₂cos(θ) + jT₂sin(θ)	T₁ = iT₁cos(φ) + jT₁sin(φ)
T = iT + j0	T = –iT + j0
W₂ = i0 – jm₂g	W₁ = i0 – jm₁g
a₂ = i0 + j0	a₁ = i0 + j0

Next we apply Newton’s Second Law separately to the i and j components for each block:

Object	i	j
	ΣF_x = ma_x	ΣF_y = ma_y
Left Block	T₂ cos(θ) - T = 0	T₂sin(θ) - m₂g = 0
Right Block	T₁ cos(φ) - T = 0	T₁sin(φ) - m₁g = 0

Considering the equations for the left block, we see that

T = T₂ cos(θ) .

We don't know T₂ however. But we do have another equation with T₂ in it and it indicates that

T₂ = m₂g / sin(θ) .

Hence the tension in the string between the blocks is

T = T₂ cos(θ) = m₂g cos(θ) / sin(θ) = 3.35(9.81)cos(60º)/sin(60º) = 18.97 N .

From the equations for the right block we see that

m₁ = T₁sin(φ)/g .

However, we don't know T₁. We do have a second equation that tells us that

T₁ = T / cos(φ) .

Combining these two results yields

m₁ = [T / cos(φ) ] sin(φ )/g = (18.97)sin(45º) / 9.81cos(45º) = 1.93 kg.

The first block has a mass of 1.93 kg.

In the diagram below, the mass of the object is 50.0 kg. What are the tensions in the ropes?

Since this problem deals with forces, the tensions, and we known that the acceleration is zero since nothing is moving, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for the block and the knot. We need a FBD for the knot since this is where the tensions all meet. A knot is massless since we are dealing with massless ropes. The block has mass, so it has weight. Each rope represents a different tension.

For each object, we break the forces and acceleration into components

*knot*	*block*
T₃ = iT₃ sin (45°) + jT₃ cos (45°)
T₂ = –iT + j0	T₁ = i0 + jT₁
T₁ = i0 – jT₁	W = i0 – jmg
a_knot = i0 + j0	a_block = i0 + j0

Applying Newton's Second law:

knot
block

ΣF_y = ma_y ΣF_y = ma_y ΣF_y = ma_y

T₂− T₃cos(45°) = 0 T₃sin(45°) − T₁ = 0 T₁ − mg = 0

From the above we see that T₁ = mg = 490.5 N, T₃ = T₁/sin(45°) = 693.7, and T₂ = T₃cos(45°) = 490.5 N.

The apparent weight of a person in an elevator is 7/8^th of his actual weight. What is the acceleration (including the direction) of the elevator?

This problem deals with a force, weight, and acceleration. That suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). Since the person has mass, he has weight. Since the person touches the floor of the elevator, there is a normal force from the floor. We will assume that the person will accelerate upwards.

Applying Newton's Second Law yields,

ΣF_y = ma_y

N − mg = ma.

This isn't enough to solve the problem, since we have one equation but two unknowns, N and a.

However, the apparent weight is just N, so the first sentence of the problem is N = (7/8)mg. Now we can find a,

a = (N − mg) / m = [(7/8)mg − mg]/m = −g/8 = −1.23 m/s².

The minus sign indicates that, contrary to our initial assumption, the elevator accelerates downwards at 1.23 m/s².

A 1200-kg elevator is carrying an 80.0-kg passenger. Calculate the acceleration of the elevator (and hence of the partner) if the tension in the cable pulling the elevator is (a) 15,000 N, (b) 12,557 N, and (c) 10,000 N. What is the apparent weight of the passenger in each case. Assume g = 9.81 m/s².

This problem deals with a force, tension, and acceleration. That suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for each object, the person and the elevator. Each has mass, so each has weight. Since the person stands on the elevator, there is a an equal but opposite normal force on each. The rope, and thus the tension, acts directly only on the elevator. Both object have the same acceleration if they remain in contact. We will assume that the acceleration is upwards.

Applying Newton's Second law:

elevator passenger

ΣF_y = ma_y ΣF_y = ma_y

T − Mg − N = +Ma N − mg = ma

So we have two equation in two unknowns, N and a. First we add the two equations together to get T − (M+m)g = (M+m)a. This our equation for the acceleration is a = T / (M+m) − g. The apparent weight, that is the normal force acting on the passenger, is given by N = mg + ma. When we plug in the numbers we find:

acceleration apparent weight

(a) 1.91 m/s² 938 N

(b) 0 785 N

(c) −2.00 m/s² 625 N

The negative acceleration in part (c) indicates that the elevator is actually accelerating downwards.

A person is standing on a weigh scale in an elevator. When the elevator is accelerating upwards with constant acceleration a the scale reads 867.0 N. When the elevator is accelerating downwards with acceleration a, the scale reads 604.5 N. Determine the acceleration a, the true weight of the person, and the mass of the person.

This problem deals with a force, the scale reading, and acceleration. That suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). Since the person has mass, he has weight. Since the person touches the scale, there is a normal force from the scale. The scale reading is a measure of this normal force. Since there are two cases, we draw an FBD for each case.

up down

ΣF_y = ma_y ΣF_y = ma_y

N₂ − mg = +ma N₂ − mg = −ma

So we have two equation in two unknowns, m and a. First we add the two equations together to get N₁ − mg + N₂ − mg = 0. This becomes m = (N₁ + N₂) / 2g = 75.0 kg.

Using the first equation, we get a = (N₂ − mg) / m = 1.75 m/s².

In the diagram below, m = 10.0 kg, θ = 25.0°, μ_s = 0.70, and μ_k = 0.50. What is the frictional force if F is (a) 25 N, (b) 65 N, and (c) 100 N? What is the acceleration in each case?

Since this problem deals with forces, the applied force and friction, and we are asked about the acceleration, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for the block. We will set up the axes along the incline. The block has mass, so it has weight. There is a normal for the block from the incline. There is friction but it is not immediately clear if it is static or kinetic.

Let's assume that the block is moving so that there is kinetic friction.

Next we break all the vectors into components:

F = iFcos(θ) + jFsin(θ),

N = i0 + jN,

f_k = –if_k + j0

W = i0 – jmg, and

a = ia + j0.

We apply Newton’s Second Law separately to the i and j components. Thus ΣF_x = ma_x yields

Fcos(θ) - f_k = ma

And ΣF_y = ma_y produces

N +Fsin(θ) - mg = 0

By definition, f_k = μ_kN.

From the equation in the second column, we have N = mg −Fsin(θ). Thus f_k = μ_k[mg − Fsin(θ)]. Putting this into the first equation yields

Fcos(θ) − μ_k[mg − Fsin(θ)] = ma .

Rearranging and solving for a yields,

a = (F/m)[cos(θ) + μ_ksin(θ)]− μ_kg .

Solving for the three cases:

F

(N)

a

(m/s²)

(a) 25 −2.11

(b) 65 +2.36

(c) 100 +6.27

Clearly, our answer to part (a) violates our assumption that the friction is kinetic. Hence the friction must have been static and a = 0 for case (a).

(a) If m = 10.0 kg, μ_k = 0.3, and F = 50 N x, find a.

(b) If m = 10.0kg, a = 2.5m/s² x and F = 75 N x, find μ_k.

Since this problem deals with forces, friction and F, and acceleration is involved, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for the block. We will set up the axes along the direction of acceleration. The block has mass, so it has weight. There is a normal for the block from the table top. Presumably the block is moving so there is kinetic friction opposed to the motion.

i

j

ΣF_x = ma_x ΣF_y = ma_y

F − f_k = ma N − mg = 0

By definition, f_k = μ_kN. From the equation in the second column, we have N = mg. Thus f_k = μ_kmg. Putting this into the first equation yields

F − μ_kmg = ma .

(a) Solving for the acceleration

a = F/m − μ_kg = (50.0 N)/(10.0 kg) − 0.3(9.81 m/s²) = 2.06 m/s² .

(b) Solving for μ_k

μ_k = F/mg − a/g = (75 n)/(10.0 kg × 9.81 m/s²)− 2.5/9.81 = 0.51 .

A 15.0-kg block sitting on a floor is pulled by two forces as shown in the diagram below. The block does not move. Find the type, magnitude, and direction of the frictional force acting on the block. Note F₁ = 48.0 N and F₂ = 42.0 N.

Since the block does not move, we know that we are dealing with static friction. It can either be to the right or to the left in the diagram. To determine which, we must examine the magnitude of each force's x component:

F_1x = F₁cos(40º) = 36.77 N,

and

F_2x = F₂cos(20º) = 39.47 N.

Since F₂x is bigger, the frictional force must be directed to the right. Note that nothing in the problem says that the object is about to slip, so we are not dealing with f_s^max. Considering the forces in the x direction

− F_2x + F_1x + f_s = 0 .

f_s = F_2x − F_1x = 39.47 − 36.77 N = 2.70 N .

A block weighing 100 N is placed on a plane with slope angle 30.0° and is pulled by a rope which is parallel to the incline and under tension T as shown below. The coefficient of static friction is 0.52, and the coefficient of kinetic friction is 0.20.
(a) Find the tension T for which the block moves up the plane at constant speed, once it has been set in motion.
(b) Find the tension T for which the block moves down the plane at constant speed, once it has been set in motion.
(c) What is the maximum value of T for which the block will remain at rest?
(d) What is the minimum value of T for which the block will remain at rest?

Since this problem deals with forces, weight and friction, and we known that the acceleration is zero since the velocity is constant or zero, that suggests we should apply Newton’s Second Law. To apply Newton’s Second Law we draw a free-body diagram (FBD) for the block. We will set up the axes along the incline. The tension in the rope points along the rope away from the block. The block has mass, so it has weight. There is a normal for the block from the incline. There is friction but it is not immediately clear if it is static or kinetic.

(a) Here the block is moving up the incline, so we have kinetic friction down the incline.

Next we break all the vectors into components:

T = iT + j0,
N = i0 + jN,
f_k = –if_k + j0
W = –iWsin(θ) – jWcos(θ), and

a = i0 + j0.

We apply Newton’s Second Law separately to the i and j components. Thus ΣF_x = ma_x yields

T - f_k - Wsin(θ) = 0

And ΣF_y = ma_y produces

N - Wcos(θ) = 0

By definition, f_k = μ_kN. From the equation in the second column, we have N = Wcos(θ ). Thus f_k = μ_kWcos(θ). Putting this into the first equation yields

T − μ_kWcos(θ)− Wsin(θ) = 0 .

Rearranging gives an expression for T,

T = W[sin(θ) + μ_kcos(θ)] = (100 N)[sin(30°)+0.20cos(30°)] = 67.3 N .

(b) Here the block is moving down the incline, so the kinetic friction is up the incline.

i j

ΣF_x = ma_x ΣF_y = ma_y

T + f_k − Wsin(θ) = 0 N − Wcos(θ) = 0

By definition, f_k = μ_kN. From the equation in the second column, we have N = Wcos(θ). Thus f_k = μ_kWcos(θ). Putting this into the first equation yields,

T + μ_kWcos(θ) − Wsin(θ) = 0 .

Rearranging gives an expression for T,

T = W[sin(θ ) − μ_kcos(θ)] = (100 N)[sin(30°)−0.20cos(30°)] = 32.7 N .

(c) The block doesn’t move, so we must be dealing with static friction. If we pull too hard the block will move up the incline, so the static friction must point down the incline. So we only need to redo (a) with static friction. Nothing will change except the coefficient of friction, so are results will be the same but with μ_k replaced by μ_s. Thus our results will be

T_max = W[sin(θ) + μ_scos(θ)] = (100 N)[sin(30°)+0.52cos(30°)] = 95.0 N ,

The block doesn’t move, so we must be dealing with static friction. If we don't pull hard enough the block will slide down the incline, so the static friction must point up the incline. So we only need to redo (b) with static friction. Nothing will change except the coefficient of friction, so are results will be the same but with μ_k replaced by μ_s. Thus our results will be

T_min = W[sin(θ)− μ_scos(θ)] = (100 N)[sin(30°)−0.52cos(30°)] = 5.0 N .

In the diagram below, block A weighs 100 N. The coefficient of static friction between the block and the surface on which it rests is 0.40. Find the maximum weight w for which the system will remain in equilibrium.

A system in equilibrium does not accelerate, so a = 0. Since this problem deals with forces, friction and weight, and we known that the acceleration is zero, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD) for each block and the knot. We need a FBD for the knot since this is where the tensions all meet. A knot is massless since we are dealing with massless ropes. Each block has mass, so each has weight. Each rope represents a different tension. Block A is on a table so there is a normal force from the table through A. If there were no friction, Block A would move forward. We deduce that the maximum static friction points left.

For each object, we break the forces and acceleration into components

*block A*	*knot*	*weight*
T₁ = i0 + jT₁	T₁ = i0 – jT₁
f = –if_sMAX + j0	T₂ = iT₂cos(40°) + jT₂sin(40°)
N = i0 + jN	T₃ = i0 – jT₃	T₃ = i0 + jT₃
W_A = i0 – jmg		W = i0 – jw
a = i0 + j0

Applying Newton’s Second law:

*Block A*		*knot*		w
i	j	i	j	j
ΣF_x = ma_x	ΣF_y = ma_y	ΣF_x = ma_x	ΣF_y = ma_y	ΣF_y = ma_y
T₁ − f_{s MAX} = 0	N − mg = 0	T₂cos(40°) − T₁ = 0	T₂sin(40°) − T₃ = 0	T₃ − w = 0

Besides our equations in the last row above, we also have the definition f_{s MAX} = μN. Our equation in the second column tells us that N = mg = 100 N, so we know f_s
MAX = μN = 0.40100 N = 40 N. The first equation tells us that T₁ = f_{s MAX} = 40 N. The third equation yields, T₂ = T₁/ cos(40°) = 52.2 N. The fourth equation yields T₃ = T₂sin(40°) = 33.6 N. The equation in the last column indicates that w = T₃ = 33.6 N. So the maximum weight of the hanging block is 33.6 N.

When a force of 500 N pushes on a 25-kg box as shown in the figure below, the acceleration of the box up the incline is 0.75 m/s². Find the coefficient of sliding friction between box and incline.

Since this problem deals with forces, F, weight, and friction, and we are given the acceleration, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). We are told that there is an applied force F acting horizontally. Since the block has mass, it has weight. It is on the incline, so there is a normal. The object is accelerating, so we must be dealing with kinetic friction which will be opposite to the direction of motion. Note that we will choose a set of axes such that one axis points along the incline.

Next we break all the vectors into components:

F = iF cos(θ) – jF sin(θ),
N = i0 + jN,
f_k = –if_k + j0
W = –imgsin(θ) – jmgcos(θ), and

a = ia + j0

We apply Newton’s Second Law separately to the i and j components. Thus ΣF_x = ma_x yields

Fcos(θ) - f_k - mgsin(θ) = ma

And ΣF_y = ma_y produces

N - mgcos(θ) - Fsin(θ) = 0

Since we know f_k = μ_kN, and the second equation says that N = mgcos(θ) + Fsin(θ), we have f_k = μ_k[mgcos(θ) + Fsin(θ)]. Substituting this into the first equation yields,

Fcos(θ) − μ_k[mgcos(θ) + Fsin(θ)] − mgsin(θ) = ma .

Isolating the term involving μ_k yields

−μ_k[mgsin(θ) + Fsin(θ)] = −Fcos(θ) + mgsin(θ) + ma .

So we find that

The kinetic coefficient of friction is 0.41 .

A 20-kg box sits on an incline as shown below. The coefficient of kinetic friction between the block and incline is 0.30. Find the acceleration of the box down the incline.

Since this problem deals with a force, friction, and we are asked about the acceleration, that suggests we should apply Newton's Second Law. To apply Newton's Second Law we draw a free-body diagram (FBD). Since the box has mass, it has weight. It is on the incline, so there is a normal. For a moving object, we must be dealing with kinetic friction. Kinetic friction opposes the motion, so it will point up the incline. Note that we should choose a set of axes such that one axis points along the acceleration as this simplifies our equations.

Next we break all the vectors into components:

N = i0 + jN,
f_k = if_k + j0
W = –imgsin(θ) – jmgcos(θ), and

a = –ia + j0.

We apply Newton’s Second Law separately to the i and j components. Thus ΣF_x = ma_x yields

f_k - mgsin(θ) = -ma

And ΣF_y = ma_y produces

N - mgcos(θ) = 0

Since we know f_k = μ_kN, and the second equation says that N = mgcos(θ), we have f_k = μ_kmgcos(θ). Substituting this into the first equation yields,

μ_kmgcos(θ) − mgsin(θ) = −ma .

Solving for a yields

a = gsin(θ) − μ_kmgcos(θ) = (9.81 m/s²)[sin(30°)-0.30cos(30°)] = 2.36 m/s².

The box accelerates down the incline at 2.36 m/s².

A 5.0-kg blog is on the a 20° incline. The coefficients of friction between the block and incline are μ_s = 0.52 and μ_k = 0.38.
(a) Does the block stay in place if the block is gently placed on the incline?
(b) If you gave the block a gentle nudge, would it slide down the incline?
(c) What must be the minimum angle θ of the incline if the block will always slide down the incline and never stay in place?

To solve these questions, we will draw the free-body diagram of the block and apply Newton’s Laws. The block has mass, so weight mg acts straight down to the bottom of the page. The block is on a surface, so there is a normal force, N, perpendicular to the surface and through the block. Finally, there is friction if the block is to be stationary. It must be f_s acting up the incline. Also since the block is stationary, a = 0.

Since the block is on an incline, the problem is simplified if we choose a tilted coordinate system. Also note that the weight mg makes an angle θ to the chosen vertical axis. This is all shown in the diagram below.

We next determine the equations in the x and y directions.

N − mgcosθ = 0

f_s − mgsinθ = 0

(a) To this question, we must evaluate and compare f_s and f_s^max = μ_sN. If f_s < f_s^max, the block does stay in place.

From the second equation f_s = mgsinθ = (5.0)(9.81)sin20° = 16.776 N. And from the first equation, N = mgcosθ = (5.0)(9.81)cos20° = 49.05 N. From this f_s^max = μ_sN = (0.52)(49.05) = 25.506 N. The block does stay in place.

(b) By nudging the block a little, we are allowing kinetic friction f_k = μ_kN to act on the block. If f_k > f_s, the block would not move. The component of weight down the incline could not overpower f_k. Evaluating f_k = μ_kN = (0.38)(49.05) = 18.639 N. The block will not slide down the incline even with a nudge.

(c) As the angle θ increases, the second of our Newton’s Equation indicates that f_s increases. But f_s cannot exceed f_s^max = μ_sN. If we replace f_s with f_s^max in the second equation, we should be able to find the critical angle at which the block would be about to slip.

f_s^max − mgsinθ = 0

μ_sN − mgsinθ = 0

μ_smgcosθ − mgsinθ = 0

mg can be cancelled out of the last equation because it is a common factor. Recalling that tanθ = sinθ / cosθ, the last result can be written as tanθ = μ_s. We then find θ = arctan(μ_s) = arctan(0.52) = 27.5°.

A 10.0-kg block is on a tabletop. The coefficients of friction between the block and tabletop are μ_s = 0.28 and μ_k = 0.18. A string is strung over a massless pulley to a hanging mass of 2.0 kg. But the hanging block is being held to keep everything motionless.
(a) Does the block on the tabletop move, if the hanging block is released?
(b) How much friction is acting on the block on the tabletop?
(c) What mass of hanging block is needed to just get the blocks to move?
(d) If there is a slight nudge given in (c), what is the acceleration of the blocks?

We start by drawing the free-body diagram of each block. Each block has mass, mg, acting straight down to the bottom of the page. Since the masses are different, we use subscripts to distinguish between them. The string exerts a tension T directed away from each block. The block on the tabletop experiences a Normal force, N, from the tabletop through the block.

We are being asked a what if question about the motion. The approach to take is to assume that there is a static friction, f_s, acting to keep the sytem stationary. Friction must act to the left for this to be true. The diagrams are shown below.

(a) To answer the question, we apply Newton's Law to the diagrams and examine the resulting equations. If the friction f_s can equal the tension T without exceeding f_s^max, there will be no motion.

The equations we find are:

N − m₁g = 0	[1]
T − f_s = 0	[2]
T − m₂g = 0	[3]

Equations [2] and [3] can be used to eliminate T to yield f_s = m₂g = (2.0)(9.81) = 19.62 N.

Equation [1] yields, N = m₁g = (10.0)(9.81) = 98.1 N. Thus f_s^max = μ_sN = (0.28)(98.1) = 27.468 N. So if the block is released gently, the system does not move.

(b) We found f_s = 19.62 N.

(c) To just get the system to move, the hanging mass would need to make the frictional force on the tabletop block reach f_s^max. That is:

f_s = m₂g

f_s^max = m₂g

27.468 = m₂g

Solving for m₂ = (27.468)/(9.81) = 2.494 kg.

(d) The slight nuudge would switch the friction to kinetic friction, f_k = μN = (0.18)(98.1) = 17.658 N and the equations would become:

N − m₁g = 0	[4]
T − f_k = m₁a	[5]
T − m₂g = −m₂a	[6]

Equation [6] yields T = m₂g − m₂a. This can then be substituted into equation [5] to yield

m₂g − m₂a − f_k = m₁a

This can be rearranged as

m₂g − f_k = (m₁ + m₂)a

Solving for a we get a = [m₂g − f_k]/(m₁ + m₂) = [(2.494)(9.81) - 17.658] / (10 + 2.494) = 0.545 m/s².

A 10-kg block is hanging from a spring with spring constant K = 1000 N/m. The spring is attached to the ceiling of an elevator. The elevator is currently moving upwards at 10 m/s and slowing down at 1.0 m/s². How much is the spring stretched?

The elevator is moving upwards but slowing so the acceleration points downwards.

After drawing the free-body diagram, we apply Newton's Second Law and get kx − mg = −ma. Solving for x we find

x = m(g − a)/K = 10(9.81 − 1.0)/1000 = 0.088 m .

The spring is stretched 8.8 cm.

A 10-kg block is sitting on a vertical spring with spring constant K = 1000 N/m. The spring is attached to the floor of an elevator. The elevator is currently moving upwards at 10 m/s and slowing down at 1.0 m/s². How much is the spring stretched or compressed? Indicate which.

The elevator is moving upwards but slowing so the acceleration points downwards.

After drawing the free-body diagram, we apply Newton's Second Law and get kx − mg = −ma. Solving for x we find

x = m(g − a)/K = 10(9.81 − 1.0)/1000 = 0.088 m .

The spring is compressed 8.8 cm.

F_net	=	i[18.432 + -12.697 + -18.579] + j[16.890 + -8.8904 + 8.664]
	=	i[-12.844] + j[16.663]

v₀	= 0
v_f	= 30 km/h (1000 m / 1 km) (1h / 3600 s) = 8.333 m/s
Δx	= 3.50 m
a	= ?

Left Block	Right Block
	F = iFcos(37°) + jFsin(37°)
T = iT + j0	T = –iT + j0
N₂ = i0 – jN₂	N₁ = i0 – jN₁
W₂ = i0 – jm₂g	W₁ = i0 – jm₁g
a₂ = ia + j0	a₁ = ia + j0

*elevator*	*passenger*
ΣF_y = ma_y	ΣF_y = ma_y
T − Mg − N = +Ma	N − mg = ma

	*acceleration*	*apparent weight*
(a)	1.91 m/s²	938 N
(b)	0	785 N
(c)	−2.00 m/s²	625 N

up	*down*
ΣF_y = ma_y	ΣF_y = ma_y
N₂ − mg = +ma	N₂ − mg = −ma

i	j
ΣF_x = ma_x	ΣF_y = ma_y
T + f_k − Wsin(θ) = 0	N − Wcos(θ) = 0

f_s^max − mgsinθ = 0
μ_sN − mgsinθ = 0
μ_smgcosθ − mgsinθ = 0