Questions: 1 2 3 4 5 6 7 8 9 10 11

1. The distance between the oxygen molecule and each of the hydrogen atoms in a water (H₂O) molecule is 0.0958 nm; the angle between the two oxygen-hydrogen bonds is 105°. Treating the atoms as particles, find the centre of mass.
   

   The problem expects you to recall that the mass of an oxygen atom is 16 times that of a hydrogen atom. The first step is to choose a coordinate system, such as the one in the diagram, and locate each particle. The chosen origin is the centre of the box.

   

   Atom	Mass (H)	xi	yi	mixi	miyi
   H	1	-0.0958sin15	0.0958cos15	-0.02479	0.09254
   O	16	0	0	0	0
   H	1	0.0958	0	0.0958	0
   Totals:	18			0.07101	0.09254

   The coordinates of the centre of mass are given by

   x_cm = (Σm_ix_i)/M_total = 0.07101/18 = 0.0039 nm, and
   y_cm = (m_iy_i)/M_total = 0.09254/18 = 0.0051 nm.

   The centre of mass is located at (0.0039 nm, 0.0051 nm). Answers will vary based on the choice of coordinate system.

   

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2. Where is the centre of mass of a uniform cubic box of side length L which has no lid?
   

   In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes is at their easy to find geometric centre if the object is uniform. Each pierce can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles. In this case, each of the five sides can be considered a separate particle.

   The next step is to choose a coordinate system, such as the one in the diagram below, and locate each particle. The origin is at the centre of the box.

   

   Side	Mass	xi	yi	zi	mixi	miyi	mizi
   bottom	M	0	0	-½L	0	0	-½ML
   front	M	0	-½L	0	0	-½ML	0
   back	M	0	½L	0	0	½ML	0
   left	M	-½L	0	0	-½ML	0	0
   right	M	½L	0	0	½ML	0	0
   Totals:	5M				0	0	-½ML

   The coordinates of the centre of mass are given by
   

   x_cm = (Σm_ix_i)/M_total = 0,
   y_cm = (Σm_iy_i)/M_total = 0, and
   z_cm = (Σm_iz_i)/M_total = -½ML / 5M = -L/10.

   The centre of mass is located at (0, 0, -L/10). Answers will vary based on the choice of coordinate system.

   It is also permissible to use symmetry arguments. For example, the figure in the diagram is only unbalanced in the z direction, thus we know that x_cm = y_cm = 0. We only needed the z columns in the above table.
   

   

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3. Two uniform squares of sheet metal of dimension L × L are joined at a right angle along one edge. One of the squares has twice the mass of the other. Find the centre of mass.
   

   In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes is easy to find, it is the geometric centre if the object is uniform. Each pierce can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles. In this case, we have one particle of mass M located in the centre of the lighter side, and a mass of 2M in the centre of the heavier side.

   The next step is to choose a coordinate system, such as the one in the diagram below, and locate each particle. The origin is in the centre of the join of the two plates.

   

   Using the symmetry of the problem, we see that the CM must be located in the xz plane, we know that y_cm = 0.

   Side	Mass	xi	zi	mixi	mizi
   side	M	0	½L	0	½ML
   bottom	2M	½L	0	ML	0
   Totals:	3M			ML	½ML

   The components of the centre of mass are given by

   x_cm = (Σm_ix_i)/M_total = ML / 3M = L/3,
   z_cm = (Σm_iz_i)/M_total = ½ML / 3M = L/6.

   The centre of mass is located at (L/3, 0, L/6). Answers will vary based on the choice of coordinate system.

   

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4. Find the centre of mass of the following signs from an origin at the lower left-hand corner of each sign. The signs are uniform and the dimensions are given.

   (a)

   (b)

   (c)

   In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes is easy to find, it is the geometric centre if the object is uniform. Each pierce can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles.

   (a)		The area and the location of CM of each rectangle with respect to the given origin (0,0) is found. Since the sign is uniform, mass can be replaced by area in the formula xCM = (Σmixi)/Mtotal. xCM = (6L2 × 0.5L + 3L2 × 2.5L)/(6L2 + 3L2) = (10.5L)/9 = (21/18)L yCM = (6L2 × 3L + 3L2 × 0.5L)/(6L2 + 3L2) = (19.5L)/9 = (39/18)L
   (b)		The area and the location of CM of each rectangle with respect to the given origin (0,0) is found. Since the sign is uniform, mass can be replaced by area in the formula xCM = (Σmixi)/Mtotal. xCM = (6L2 × 0.5L + 3L2 × 2.5L + 3L2 × 2.5)/(6L2 + 3L2 + 3L2) = (18L)/12 = (3/2)L The shape is symmetric vertically about a line drawn horizontally through the middle, so by symmetry yCM = 3L. However, we can check if desired that we get the same answer using the formula. yCM = (6L2 × 3L + 3L2 × 5.5L + 3L2 × 0.5L)/(6L2 + 3L2 + 3L2) = (36L)/12 = 3L.
   		Alternate decomposition. We should always aim for the biggest symmetric shapes that we see. In this diagram, we have treated the two horizontal rectangles as one object. The area and the location of CM of the two pieces with respect to the given origin (0,0) is found. This won't change our answer, but will be a shorter calculation. Since the sign is uniform, mass can be replaced by area in the formula xCM = (Σmixi)/Mtotal. xCM = (6L2 × 0.5L + 6L2 × 2.5)/(6L2 + 6L2) = (18L)/12 = (3/2)L The vertical symmetry is even more obvious with this decomposition, so by symmetry yCM = 3L. However, we can check if desired that we get the same answer using the formula. yCM = (6L2 × 3L + 6L2 × 3L)/(6L2 + 6L2) = (36L)/12 = 3L.
   (c)		The P shape has been split into a vertical rectangle and a hollow square. The area and the location of CM of each piece with respect to the given origin (0,0) is found. Since the sign is uniform, mass can be replaced by area in the formula xCM = (Σmixi)/Mtotal. xCM = (4L2 × 0.5L + 12L2 × 2L)/(4L2 + 12L2) = (26L)/16 = (13/8)L yCM = (4L2 × 2L + 12L2 × 6L)/(4L2 + 12L2) = (80L)/16 = 5L



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5. An L-shaped object of uniform density is hung over a nail so that it is free to pivot. What angle, θ, does the long side make with the vertical? The long side of the L-shaped object is twice as long as the short side?
   

   The problem mentioned that the object is free to pivot, to rotate. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.

   The forces that we know are working on the L-shaped object are a normal from the nail and the weight which acts from the centre of mass. Ordinarily, for complex shapes, we first determine the CM. However, in this case, it is easier to consider the two arms of the objects as being separate objects. The long arm will have a mass (2/3)mg and the short arm will be (1/3)mg. We do not have a simple method of figuring out which way the normal points. As with all pins, we consider it as two forces one vertical and one horizontal.

   

   ΣFx = 0	ΣFy = 0
   Nx = 0	Ny − (1/3)mg − (2/3)mg = 0

   These tell us the obvious, the normal has no horizontal component and that it supports the weight of the object.
   

   We will use Method A for the torques since that method is easiest to apply here. We will take the nail as the pivot point since this eliminates the torques from the nail.

   r	F		direction	τz = rFsinθ
   0	Nx	-	-	0
   0	Ny	-	-	0
   L/2	(1/3)mg	π+θ	CW	-mgLsin(π+θ)/6
   L	(2/3)mg	θ	CCW	2mgLsinθ/3

   Since Στ_z = 0, the equation we get is

   -mgLsin(π+θ)/6 + 2mgLsinθ/3 = 0.

   Eliminating common terms and noting sin(π+θ) = cosθ, this becomes

   -cosθ/2 + 2sinθ = 0,

   or

   sinθ/cosθ = 1/4.

   Using the identity, tanθ = sinθ/cosθ, we thus have θ = tan^-1(1/4) = 14.0°. The long side makes a 14.0° angle with the vertical.
   

   

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6. A uniform 400 N boom is supported as shown in the figure below. Find the tension in the tie rope and the force exerted on the boom by the pin at P.
   
   

   The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.
   

   The forces that we know are working on the boom are a normal from the pin, the weight which acts from the centre of mass, and the two tensions. We are given T₂. The CM is obviously at the centre of the boom. We do not have a simple method of figuring out which way the normal points, instead we consider it as two forces one vertical and one horizontal.

   

   ΣFx = 0	ΣFy = 0
   Px − T1 = 0	Py − mg − T2 = 0

   These tell us the obvious, that P_x = T₁ and P_y = mg + T₂ = 2400 N.
   

   We will use Method A for the torques since that method is easiest to apply here since the distances and angles are relatively easy to find. We will take the pin as the pivot point since this eliminates the torques from the pin.

   r	F	θ	direction	τz = rFsinθ
   0	Px	-	-	0
   0	Py	-	-	0
   L/2	W	40°	CW	-LWsin(40°)/2
   (3/4)L	T1	50°	CCW	3LT1sin(50°)/4
   L	T2	40°	CW	-LT2sin(40°)

   Since Στ_z = 0, the equation we get is

   -WLsin(40°)/2 + 3LT₁sin(50°)/4 - LT₂sin(40°) = 0 .

   Eliminating L from the above and rearranging to get T₁ by itself yields,
   

   T₁ = {2[W + 2T₂]sin(40°)} / 3sin(50°) .

   Using the values we are given, we find T₁ = 2461 N.

   Since we know P_x = T₁, we also know the pin force is
   

   P = i2461 N + j2400 N.

   The magnitude of this force is P = [(P_x)² + (P_y)² ]^½ = 3438 N. The force is directed at an angle θ = tan^-1(P_y/P_x) = 44.3° to the horizontal. Note that the pin force is not pointed solely along the length of the boom as one might expect.
   

   Big Point To Remember: Pin forces are not always directed in the obvious direction.

   

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7. In the figure below, a mass of 500 kg is held motionless in the air by a 120-kg boom and a rope. Find the tension in the rope. Find the force exerted on the boom by the pin at P. The angles are θ = 30.0° and φ = 45.0°.
   

   The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.
   

   The forces that we know are working on the boom are a normal from the pin, the weight which acts from the centre of mass, and the two tensions. The CM is obviously at the centre of the boom. We do not have a simple method of figuring out which way the normal points, instead we consider it as two forces one vertical and one horizontal.

   

   Fx = 0	ΣFy = 0
   Px − T1cos(π/2-θ) = 0	Py − mg − T2 − T1sinθ = 0

   These tell us that P_x = T₁cosθ and P_y = mg + T₂ + T₁sinθ. We are given the mass of the load so we know T₂ = m_loadg = 4905 N.

   We will use Method A for the torques since that method is easiest to apply here since the distances and angles are relatively easy to find. We will take the pin as the pivot point since this eliminates the torques from the pin.

   r	F	θ	direction	τz = rFsinθ
   0	Px	-	-	0
   0	Py	-	-	0
   L/2	mg	π/2 − φ	CW	-Lmgsin(π/2-φ)/2
   L	T1	θ − φ	CCW	LT1sin(θ-φ)
   L	T2	π/2 − φ	CW	-LT2sin(π/2-φ)

   Since Στ_z = 0, the equation we get is
   

   -Lmg[sin(π/2-φ)]/2 + LT₁sin(θ-φ) - LT₂sin(π/2-φ) = 0 .

   Eliminating L from the above, multiplying through by 2, and using the identity that sin(π/2-φ) = cosφ yields,
   

   -mgcosφ + 2T₁sin(θ-φ) - 2T₂cosφ = 0.

   We rearrange to get T₁ by itself,
   

   T₁ = (mg + 2T₂)cosθ / 2sin(θ-φ) .

   Using the values we are given, and the value for T₂, we find T₁ = 15008 N.

   We have P_x = T₁cosθ = 12998 N. As well, P_y = mg + T₂ + T₁sinθ = 13587 N. Thus we also know that the pin force is
   

   P = i12998 N + j13587 N.

   The magnitude of this force is P = [(P_x)² + (P_y)² ]^½ = 1.88 × 10⁴ N. The force is directed at an angle θ = tan^-1(P_y/P_x) = 46.3° to the horizontal. Note that the pin force is not pointed solely along the length of the boom as one might expect.
   

   

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8. A rectangular sign of mass 50.0 kg and width w = 5.00 m and l = length 4.00 m is hanging from a hinge and a rope as shown in the figure below. The rope makes and angle θ = 65.0° with the right wall.
   (a) Find the tension in the rope.
   (b) Find the horizontal and vertical components of the hinge force.
   
   

   The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.
   

   The forces that we know are working on the sign are a normal from the pin, the weight which acts from the centre of mass, and the tension. The CM is obviously at the centre of the sign. We do not have a simple method of figuring out which way the normal points, instead we consider it as two forces one vertical and one horizontal.

   

   ΣFx = 0	ΣFy = 0
   Px − Tsinθ = 0	Py − mg + Tcosθ = 0

   These tell us that P_x = Tsinθ and P_y = mg − Tcosθ.
   

   We will use Method B for the torques since that method is easiest to apply here since the location of the forces easy to find. We will take the pin as the pivot point since this eliminates the torques from the pin.

   x	y	Fx	Fy	τz = xFy − yFx
   0	0	Px	Py	0
   w	0	Tsinθ	Tcosθ	wTcosθ
   w/2	-l/2	0	-mg	-wmg/2

   Since Στ_z = 0, the equation we get is
   

   wTcosθ − wmg/2 = 0 .

   Eliminating w from the above and rearranging yields,
   

   T = mg / 2cosθ = 580.3 N .

   The force equations give P_x = Tsinθ = 526 N and P_y = mg − Tcosτ_q = 245.0 N.
   

   Thus the tension in the rope is 580 N and the horizontal and vertical components of the pin force are 526 N and 245 N respectively.

   

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9. Find the centre of mass of the sign. Determine the tension in the strings and the unknown angle θ. Each square has a side of length 32.0 cm. The object has a mass of 125 g.
   
   

   In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes are easy to find, it is the geometric centre if the object is uniform. Each piece can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles.

   The area and the location of CM of each rectangle with respect to the given origin (0,0) is found. Since the sign is uniform, mass can be replaced by area in the formula xCM = (Σmixi)/Mtotal. The horizontal component of the centre of mass of the sign must be on the vertical symmetry axis. However, we will proceed with the calculation just to check. xCM = (L2 × 0.5L + L2 × 1.5L + L2 × 2.5L)/(L2 + L2 + L2) = (4.5L)/3 = 1.5L yCM = (L2 × 0.5L + L2 × −0.5L + L2 × −0.5L)/(L2 + L2 + L2) = (−0.5L)/3 = −L/6

   The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.
   

   The forces that we know are working on the sign are the two tensions and the weight which acts from the centre of mass of each piece. Each piece has one-third of the total mass and weight.

   

   ΣFx = 0	ΣFy = 0
   T1sinθ − T2sin(65°) = 0	T1cosθ + T2cos(65°) − Mg = 0

   These tell us that T₁sinθ = T₂sin(65) and T₁cosθ + T₂cos(65°) = Mg.
   

   We will use Method B for the torques since that method is easiest to apply here since the location of the forces easy to find. We will locate the pivot at the upper right corner because we have two unknowns there.

   x(m)	y( m)	Fx	Fy	τz = xFy − yFx
   0	0	T1sinθ	T1cosθ	0
   -0.64	-0.32	-T2sin(65°)	T2cos(65°)	0.64T2(65°) - 0.32T2sin(65°)
   −0.16	−0.16/6	0	−Mg	−(0.16)Mg

   Since Στ_z = 0, the equation we get is
   

   -0.64T₂cos(65°) − 0.32T₂sin(65°) + 0.16Mg = 0 .

   Rearranging the above yields the tension in the left string,

   T₂ = 0.16Mg / [0.64cos(65°) + 0.32sin(65°)] = 0.3500 N .

   The force equations give
   

   T₁sinθ = T₂sin(65°) = 0.31725 N, and
   
   T₁cosτ = Mg − T₂cos(65°) = 1.07831 N.
   

   Taking the ratio of these two results we have sinθ/cosθ = 0.31725/1.07831 or tanθ = 0.2942. So the unknown angle is θ = 16.4°. Substituting the angle back into either of the two equations yields the tension in the right string T₁ = 1.124 N.
   

   

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10. The sign has a mass of 20.0 kg. The hinge is located at the bottom of the left side. Find the centre of mass. Determine the tension in the rope and the horizontal and vertical components of the hinge force. The length, l, is 12 cm.
    

    In dealing with real objects rather than particles, we treat the complex object as a grouping of simpler shapes. The CM of the simpler shapes are easy to find, it is the geometric centre if the object is uniform. Each piece can now be considered a particle with the mass of the piece located at the CM of that piece. We have, in effect, turned the complex shape into a collection of particles.

    The area and the location of CM of each rectangle with respect to the given origin (0,0) is found. Since the sign is uniform, mass can be replaced by area in the formula xCM = (Σmixi)/Mtotal. The horizontal component of the centre of mass of the sign must be on the vertical symmetry axis. However, we will proceed with the calculation just to check. xCM = (5L2 × 2.5L + 4L2 × 2.5L)/(5L2 + 4L2) = 2.5L yCM = (5L2 × 0.5L + 4L2 × −2L)/(5L2 + 4L2) = (−5.5L)/9 = −(11/9)L

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.
    

    The forces that we know are working on the sign are the tension, and the weight which acts from the centre of mass, and the normal force from the hinge. Since we do not know the direction of the normal force, we show components.
    

    

    ΣFx = 0	ΣFy = 0
    -Hx + Tsin(40°) = 0	Hy + Tcos(40°) − Mg = 0

    These tell us that H_x = Tsin(40°) and H_y + Tcos(40°) = Mg.
    

    We will use Method B for the torques since that method is easiest to apply here since the location of the forces easy to find. We will locate the pivot at the hinge because we have two unknowns there.

    x	y	Fx	Fy	τz = xFy − yFx
    0	0	Hx	Hy	0
    5ℓ	ℓ	Tsin(40°)	Tcos(40°)	5ℓTcos(40°) − ℓTsin(40°)
    2.5ℓ	−ℓ	0	−Mg	-2.5Mgℓ

    Since Στ_z = 0, the equation we get is
    

    5ℓTcos(40°) − ℓTsin(40°) − 2.5ℓMg = 0 .

    Eliminating ℓ and rearranging the above yields the tension in the rope,
    

    T = 2.5Mg / [5cos(40°) − sin(40°)] = 153.9 N .

    The force equations give
    

    H_x = Tsin(40°) = 98.9 N , and
    H_y = Mg − Tcos(40°) = 78.3 N.

    

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11. A ladder is propped against a wall making an angle with the floor. The wall is frictionless but the coefficients of friction for the floor are μ_s and μ_k respectively. Obtain an expression for the smallest that can be if the ladder is not to slip. Recall that tanθ = sinθ/cosθ.
    
    

    The problem mentions forces and looking at the diagram shows that the object would rotate in the absence of any one of these forces. This indicates that we are dealing with a Static Equilibrium problem. We solve Static Equilibrium problems by sketching the extended free-body diagram, an FBD where the location of the all forces are indicated so that torques can be calculated. Then we determine the three equations necessary for static equilibrium, ΣF_x = 0, ΣF_y = 0, and Στ_z = 0.
    

    The forces that we know are working on the ladder are the weight which acts from the centre of mass, the normal forces from the wall and floor, and friction. Since the ladder is not moving, we are dealing with static friction. Since we want the smallest angle, we are dealing with f_{s MAX}. Since the ladder has a tendency to move to the left, friction points to the left.

    

    ΣFx = 0	ΣFy = 0
    fs MAX − Nw = 0	Nf − mg = 0

    These tell us that f_{s MAX} = μN_w and N_f = mg. We also know that f_{s MAX} = μ_sN_f where N_f is the normal force between the ladder and floor. As a result, we have f_{s MAX} = μ_smg. Hence N_w = μ_smg as well.
    

    We will use Method A for the torques since that method is easiest to apply here since the distances and angles are easy to find. We will locate the pivot at the floor because we have two unknowns there.

    r (m)	F (N)	θ	direction	τz = rFsinθ
    0	fs MAX	-	-	0
    0	Nf	-	-	0
    ½L	mg	π/2-θ	CW	-½Lmgsin(π/2-θ)
    L	Nw	θ	CCW	LNwsinθ

    Since τ_z = 0, the equation we get is
    

    -½Lmgsin(π/2-θ) + LN_wsinθ = 0 .

    Eliminating L and noting that sin(π/2-θ) = cosθ yields,
    

    -½mgcosθ + N_wsinθ = 0 .

    The force equations gave N_w = μ_smg, so we have
    

    -½mgcosθ + μ_smgsinθ = 0 .

    Rearranging and using tanθ = sinθ /cosτ, we get
    

    θ = tan^-1(1 / 2μ_s) .

    If the angle were any smaller than this, the ladder would slip.

    

Questions? mikec@kpu.ca