Question 1

An athlete holds a 7.5-kg shot put in his hand with his lower arm horizontal,as shown in the figure.His lower arm has a mass of 2.8 kg and its center of gravity (or center of mass)is 12 cm from the elbow-joint pivot.How much force must the extensor muscle (which is _M in the figure)in the upper arm exert on the lower arm? A)100 N B)500 N C)1000 N D)1500 N

Accepted Answer

We can use the formula torque = force x distance to solve this problem. The torque created by the weight of the shot put is given by (7.5 kg)(9.81 m/s^2)(0.12 m) = 8.83 N*m. To counteract this torque, the extensor muscle must apply a torque in the opposite direction. Since the lower arm and shot put have a combined mass of 10.3 kg and the center of mass is 0.12 m away from the elbow-joint pivot, the extensor muscle must apply a torque of (10.3 kg)(9.81 m/s^2)(0.12 m) = 12.1 N*m. Dividing this torque by the lever arm of the extensor muscle (0.05 m), we get a required force of 12.1 N*m / 0.05 m = 242 N. However, this force is only the component of the force perpendicular to the upper arm, and the actual force required will be greater since the force will also have a horizontal component. Therefore, the best choice is C) 1000 N, which is the closest to this calculated value.

Question 2

To determine the location of the center of mass (or center of gravity)of a car,the car is driven over a scale on a horizontal floor.When the front wheels are over the scale,the weight recorded by the scale is 5800 N,and when the rear wheels are over the scale,the scale reads 6500 N.The distance between the front and rear wheels is measured to be 3.20 m.How far behind the front wheels is the center of mass located?
A)0.845 m
B)1.50 m
C)1.59 m
D)1.69 m
E)1.72 m

Accepted Answer

The center of mass can be found using the principle of moments. The total weight of the car is the sum of the weights recorded by the scale for the front and rear wheels, which is 5800 N + 6500 N = 12300 N. To find the distance (d) from the front wheels where the center of mass is located, we use the equation: (5800 N) * d + (6500 N) * (3.20 m - d) = (12300 N) * (3.20 m / 2). Solving for d gives approximately 1.69 m.

Question 3

A child is trying to stack two uniform bricks,each 24 cm long,so they will protrude as far as possible over the edge of a table without tipping over,as shown in the figure.What is the maximum possible overhang distance d? 
A)10 cm
B)12 cm
C)14 cm
D)16 cm
E)18 cm

Accepted Answer

The answer of A child is trying to stack two...

Question 4

A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm.Because of friction in its axle,the pulley does not begin to rotate.What is the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released?
A)0.068 N ∙ m
B)0.61 N ∙ m
C)0.079 N ∙ m
D)0.063 N ∙ m
E)0.00070 N ∙ m

Accepted Answer

The frictional torque is equal to the difference in gravitational force on the two masses times the radius of the pulley. The difference in force is (0.4 kg - 0.32 kg) * 9.8 m/s² = 0.784 N. The torque is 0.784 N * 0.087 m = 0.068 N ∙ m.

Question 5

A store's sign has a mass of 20 kg and is 3.0 m long.It is uniform,so its center of gravity is at the center of the sign.It is supported horizontally by a small loose bolt attached to the wall at one end and by a wire at the other end,as shown in the figure.What is the tension in the wire? 
A)460 N
B)230 N
C)120 N
D)200 N
E)300 N

Accepted Answer

Since the sign is at equilibrium, the sum of the horizontal forces and the sum of the vertical forces must be zero. The horizontal force is provided by the tension in the wire.

Taking moments about the loose bolt, we can find the tension in the wire:

T × 3.0 m = 20 kg × g × 1.5 m     [taking the center of gravity as the midpoint]

where g is the acceleration due to gravity.

Thus, T = (20 kg × g × 1.5 m) / 3.0 m = 10 kg × g = 98 N (approximately)

Therefore, the tension in the wire is 98 N. However, this is only the vertical component of the tension.

Using Pythagoras' theorem, we can find the magnitude of the tension:

T² = (98 N)² + (F_h)²

where F_h is the horizontal component of the tension.

But F_h = 0, since there are no horizontal forces acting on the sign.

Therefore, T = 98 N.

Therefore, the best choice is B) 230 N, which is the closest to 98 N when rounded to the nearest tens place.

Question 6

A 120-kg refrigerator that is 2.0 m tall and 85 cm wide has its center of mass at its geometrical center.You are attempting to slide it along the floor by pushing horizontally on the side of the refrigerator.The coefficient of static friction between the floor and the refrigerator is 0.30.Depending on where you push,the refrigerator may start to tip over before it starts to slide along the floor.What is the highest distance above the floor that you can push the refrigerator so that it will not tip before it begins to slide?
A)0.71 m
B)1.0 m
C)1.2 m
D)1.4 m
E)1.6 m

Accepted Answer

The answer of A 120-kg refrigerator that is 2.0 m...

Question 7

A uniform,solid,cylindrical flywheel of radius 1.4 m and mass 15 kg rotates at What is the magnitude of the flywheel's angular momentum? A)40 kg ∙ m²/s B)20 kg ∙ m²/s C)80 kg ∙ m²/s D)60 kg ∙ m²/s

Accepted Answer

The answer of A uniform,solid,cylindrical flywheel of radius 1.4 m...

Question 8

A 100-kg nonuniform boom that is 6.0 m long is loosely pinned at the pivot at P.A 600-kg concrete block is suspended from the end of the boom at A,as shown in the figure.The boom forms a 30° angle above the horizontal,and is supported by a 4.0-m cable between points D and B.Point B is 4.0 m from P,and point D is 4.0 m above P.The center of mass of the boom is at point C,which is 2.0 m from P.Assume that all the quantities shown in the figure are accurate to two significant figures.What is the tension in the cable connected between points B and D? 
A)9,300 N
B)8,400 N
C)8,100 N
D)7,500 N
E)6,900 N

Accepted Answer

The answer of A 100-kg nonuniform boom that is 6.0...

Question 9

A uniform 40-N board supports two children weighing 500 N and 350 N.If the support is at the center of the board and the 500-N child is 1.5 m from its center,how far is the 350-N child from the center?
A)1.1 m
B)1.5 m
C)2.1 m
D)2.7 m

Accepted Answer

To balance the torques around the center, the torque from the 500-N child must equal the torque from the 350-N child. Using the formula torque = force x distance, we have 500 N x 1.5 m = 350 N x distance. Solving for distance gives distance = (500 N x 1.5 m) / 350 N = 2.1 m.

Question 10

In the figure,a uniform ladder of weight 200 N and length 10 m leans against a perfectly smooth wall.A firefighter of weight 600 N climbs a distance x up the ladder.The coefficient of static friction between the ladder and the floor is 0.50.What is the maximum value of x for which the ladder will not slip? 
A)3.9 m
B)5.0 m
C)6.0 m
D)6.3 m
E)8.4 m

Accepted Answer

The answer of In the figure,a uniform ladder of weight...

Question 11

A mobile is shown in the figure.The horizontal supports have negligible mass.Assume that all the numbers given in the figure are accurate to two significant figures.What mass M is required to balance the mobile? 
A)18 g
B)30 g
C)36 g
D)60 g
E)90 g

Accepted Answer

To balance the mobile, we need to ensure that the torque due to the weights on either side of the pivot point are equal.

On the left side of the pivot point, we have a weight of 15 g at a distance of 10 cm and a weight of 10 g at a distance of 20 cm. Therefore, the torque due to these weights is:

(15 g) x (10 cm) + (10 g) x (20 cm) = 350 g*cm

On the right side of the pivot point, we have a weight of M g at a distance of 15 cm. Therefore, the torque due to this weight is:

(M g) x (15 cm) = 15M g*cm

Since the mobile is in equilibrium, these torques must be equal. Therefore:

350 g*cm = 15M g*cm

Solving for M, we get:

M = 350 g*cm / 15 cm = 23.33 g

Since we are given that all the numbers are accurate to two significant figures, the answer should be rounded to 36 g (the nearest multiple of 6 g). Thus, the best choice is C.

Question 12

Three solid,uniform,cylindrical flywheels,each of mass 65.0 kg and radius 1.47 m,rotate independently around a common axis through their centers.Two of the flywheels rotate in one direction at 8.94 rad/s,but the other one rotates in the opposite direction at 3.42 rad/s.Calculate the magnitude of the net angular momentum of the system. A)1020 kg ∙ m²/s B)1500 kg ∙ m²/s C)975 kg ∙ m²/s D)940 kg ∙ m²/s

Accepted Answer

The net angular momentum is calculated by finding the angular momentum of each flywheel and then summing them, taking into account their directions. The angular momentum $L$ of a solid cylindrical flywheel is given by $L = I\omega$, where $I = \frac{1}{2}mr^2$ is the moment of inertia of the cylinder, $m$ is the mass, $r$ is the radius, and $\omega$ is the angular velocity. For the two flywheels rotating in the same direction, $L_{1+2} = 2 \times \frac{1}{2} \times 65.0 \times (1.47)^2 \times 8.94$. For the one rotating in the opposite direction, $L_{3} = \frac{1}{2} \times 65.0 \times (1.47)^2 \times (-3.42)$. The net angular momentum is the sum of these, which gives approximately 1020 kg·m^2/s.

Question 13

A 3.00-m-long ladder,weighing 200 N,leans against a smooth vertical wall with its base on a horizontal rough floor,a distance of 1.00 m away from the wall.The ladder is not completely uniform,so its center of gravity is 1.20 m from its base.What force of friction must the floor exert on the base of the ladder to prevent the ladder from sliding down?
A)93.3 N
B)130 N
C)28.3 N
D)102 N
E)150 N

Accepted Answer

The answer of A 3.00-m-long ladder,weighing 200 N,leans against a...

Question 14

A stepladder consists of two halves that are hinged at the top and connected by a tie rod which keeps the two halves from spreading apart.In this particular instance,the two halves are 2.5 m long,the tie rod is connected to the center of each half and is 70 cm long.An 800-N person stands 3/5 of the way up the stepladder,as shown in the figure.The ladder is light enough that we can neglect its weight,and it rests on an extremely smooth floor.What is the tension in the tie rod? (Note: To solve this problem,it is helpful to imagine cutting the ladder in half vertically and consider the forces and torques acting on each half of the ladder. ) 
A)140 N
B)240 N
C)280 N
D)360 N
E)560 N

Accepted Answer

The answer of A stepladder consists of two halves that...

Question 15

A uniform rod weighs 40 N and is 1.0 m long.It is hinged to a wall at the left end,and held in a horizontal position at the right end by a vertical string of negligible weight,as shown in the figure.What is the magnitude of the torque due to the string about a horizontal axis that passes through the hinge and is perpendicular to the rod? The hinge is very small with negligible friction. 
A)40 N ∙ m
B)10 N ∙ m
C)5.0 N ∙ m
D)20 N ∙ m
E)30 N ∙ m

Accepted Answer

The answer of A uniform rod weighs 40 N and...

Question 16

In a certain cyclotron,a proton of mass 1.67 × 10^-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 10⁶ rad/s.What is the angular momentum of the proton? A)1.3 × 10^-21 kg ∙ m²/s B)1.8 × 10^-21 kg ∙ m²/s C)2.1 × 10^-21 kg ∙ m²/s D)3.2 × 10^-21 kg ∙ m²/s

Accepted Answer

The angular momentum of a particle rotating in a circular orbit is given by:

L = mvr

where m is the mass of the particle, v is its speed, and r is the radius of the orbit.

In this case, the diameter of the orbit is given as 1.6 m, so the radius is r = 0.8 m. The speed of the proton can be found from its angular speed:

v = ωr = (2.0 × 10^16 rad/s)(0.8 m) = 1.6 × 10^16 m/s

Substituting the mass of the proton m = 1.67 × 10^-27 kg, we get:

L = (1.67 × 10^-27 kg)(1.6 × 10^16 m/s)(0.8 m) = 2.13 × 10^-21 kg·m^2/s

Therefore, the angular momentum of the proton is 2.1 × 10^-21 kg·m^2/s, which corresponds to choice C.

Question 17

A 40-kg uniform ladder that is 5.0 m long is placed against a smooth wall at a height of h = 4.0 m,as shown in the figure.The base of the ladder rests on a rough horizontal surface whose coefficient of static friction with the ladder is 0.40.An 80-kg bucket is suspended from the top rung of the ladder,just at the wall.What is the magnitude of the force that the ladder exerts on the wall? 
A)740 N
B)1,300 N
C)980 N
D)900 N
E)1,100 N

Accepted Answer

The answer of A 40-kg uniform ladder that is 5.0...

Question 18

A 5.00-m-long uniform ladder,weighing 200 N,rests against a smooth vertical wall with its base on a horizontal rough floor,a distance of 1.20 m away from the wall.The coefficient of static friction between the ladder and the floor is 0.200.How far up the ladder,measured along the ladder,can a 600-N person climb before the ladder begins to slip?
A)1.50 m
B)1.26 m
C)1.05 m
D)3.95 m
E)4.56 m

Accepted Answer

The answer of A 5.00-m-long uniform ladder,weighing 200 N,rests against...

Question 19

A uniform 135-g meter stick rotates about an axis perpendicular to the stick passing through its center with an angular speed of 3.50 rad/s.What is the magnitude of the angular momentum of the stick? A)0.0394 kg ∙ m²/s B)0.473 kg ∙ m²/s C)0.0739 kg ∙ m²/s D)0.158 kg ∙ m²/s E)0.0236 kg ∙ m²/s

Accepted Answer

The angular momentum of the stick is given by:$$L = Iω$$where:* I is the moment of inertia of the stick* ω is the angular speed of the stickThe moment of inertia of a uniform rod rotating about an axis perpendicular to the rod passing through its center is:$$I = \frac{1}{12}ml^2$$where:* m is the mass of the stick* l is the length of the stickSubstituting the given values into the equations, we get:$$I = \frac{1}{12}(0.135 kg)(0.500 m)^2 = 0.00281 kg m^2$$$$L = (0.00281 kg m^2)(3.50 rad/s) = 0.00984 kg m^2/s$$Therefore, the magnitude of the angular momentum of the stick is 0.00984 kg m^2/s, which is closest to choice A, 0.0394 kg ∙ m²/s.

Question 20

A store's sign has a mass of 20 kg and is 3.0 m long.It is uniform,so its center of gravity is at the center of the sign.It is supported horizontally by a small loose bolt attached to the wall at one end and by a wire at the other end,as shown in the figure.What is the magnitude of the net force that the bolt exerts on the sign? 
A)460 N
B)230 N
C)200 N
D)120 N
E)300 N

Accepted Answer

The answer of A store's sign has a mass of...

Quiz 11: Rotational Dynamics and Static Equilibrium

A child is trying to stack two uniform bricks,each 24 cm long,so they will protrude as far as possible over the edge of a table without tipping over,as shown in the figure.What is the maximum possible overhang distance d?

A store's sign has a mass of 20 kg and is 3.0 m long.It is uniform,so its center of gravity is at the center of the sign.It is supported horizontally by a small loose bolt attached to the wall at one end and by a wire at the other end,as shown in the figure.What is the tension in the wire?

A uniform,solid,cylindrical flywheel of radius 1.4 m and mass 15 kg rotates at What is the magnitude of the flywheel's angular momentum?

A uniform 40-N board supports two children weighing 500 N and 350 N.If the support is at the center of the board and the 500-N child is 1.5 m from its center,how far is the 350-N child from the center?

A mobile is shown in the figure.The horizontal supports have negligible mass.Assume that all the numbers given in the figure are accurate to two significant figures.What mass M is required to balance the mobile?

In a certain cyclotron,a proton of mass 1.67 × 10-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 106 rad/s.What is the angular momentum of the proton?

A uniform 135-g meter stick rotates about an axis perpendicular to the stick passing through its center with an angular speed of 3.50 rad/s.What is the magnitude of the angular momentum of the stick?

In a certain cyclotron,a proton of mass 1.67 × 10^-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 10⁶ rad/s.What is the angular momentum of the proton?