Question 1

A bucket filled with water has a mass of 23 kg and is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well.What torque does the weight of water and bucket produce on the cylinder if the cylinder is not permitted to rotate? (g = 9.8 m/s²) A)34 N*m B)17 N*m C)11 N*m D)23 N*m

Accepted Answer

C

Question 2

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end.How far would the worker stand from one of the supporting ropes to produce a tension of 550 N in that rope?
A)1.4 m
B)2.0 m
C)2.5 m
D)2.7 m

Accepted Answer

Let the distance of the worker from one of the supporting ropes be x. 
Taking moments about the rope, we get: 
(800N-500N) x = 550N (4.0m - x) 
Solving for x gives: 
x ≈ 2.5 m

Question 3

A meter stick is supported by a knife-edge at the 50-cm mark.Doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks, respectively.Where should Doug hang a third mass of 0.30 kg to keep the stick balanced?
A)20 cm
B)70 cm
C)30 cm
D)25 cm

Accepted Answer

C

Question 4

A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end.A 2.0-N weight is suspended from the 10-cm position on the stick, another 2.0-N weight is suspended from the 50 cm position, and a 3.0-N weight is suspended from the 60 cm position.What is the tension in the string attached at the 100-cm end of the stick?
A)1.9 N
B)3.0 N
C)3.5 N
D)4.0 N

Accepted Answer

Since the meter stick is in equilibrium, the sum of the torques about any point must be zero. We can choose the left end of the meter stick as our pivot point. Then the torques due to the weights hanging from the meter stick are:

- for 2.0-N weight at 10 cm: (0.10 m) (2.0 N) = 0.20 Nm
- for 2.0-N weight at 50 cm: (0.50 m) (2.0 N) = 1.00 Nm
- for 3.0-N weight at 60 cm: (0.60 m) (3.0 N) = 1.80 Nm

The torque due to the tension in the string at the 100-cm end is:

- for tension in string at 100 cm: (1.00 m) (T) = 1.00 T Nm

Since the meter stick is in equilibrium, the sum of the torques must be zero:

0.20 Nm + 1.00 Nm + 1.80 Nm + 1.00 T Nm = 0

Solving for T, we get:

T = -3.00 Nm / 1.00 m = -3.00 N

Since tension must be a positive quantity, we take the magnitude of the tension:

|T| = 3.00 N

Thus, the tension in the string at the 100-cm end of the stick is 3.0 N. Answer choice C is correct.

Question 5

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall.The coefficient of static friction between ladder and floor is 0.40.What minimum angle can the ladder make with the floor before it slips?
A)22$\degree$
B)51$\degree$
C)18$\degree$
D)42$\degree$

Accepted Answer

The ladder will slip if the force of friction between the ladder and the floor is less than the force component parallel to the floor. 
Let's consider the forces acting on the ladder:
- Weight (W) of the ladder acts downward, with a magnitude of 100 N.
- The normal force (N) acts upward, perpendicular to the floor, and has a magnitude of W = 100 N.
- The force of friction (f) acts parallel to the floor, opposing the ladder's tendency to slip, and has a maximum magnitude of 0.4N = 40 N (where N is the normal force).
- The ladder is in equilibrium, so the net force in the vertical (y) direction is zero: N - W*cosθ = 0, where θ is the angle the ladder makes with the floor.
- The net force in the horizontal (x) direction is also zero: f - W*sinθ = 0.

Solving these equations for θ, we get:
tanθ = f/N = W*sinθ/N*cosθ
tanθ = 0.4
θ = 51.3°

Therefore, the minimum angle the ladder can make with the floor before it slips is 51.3°. The answer is choice B.

Question 6

A 3.0-m rod is pivoted about its left end.A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a ccw torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the pivot.The 5.2 N is at an angle of 30$\degree$to the rod and causes a cw torque.What is the net torque about the pivot?
A)15 N*m
B)0 N*m
C)-6.3 N*m
D)-0.6 N*m

Accepted Answer

The answer of A 3.0-m rod is pivoted about its...

Question 7

A hoop of radius 1.0 m is placed in the first quadrant of an xy-coordinate system with its rim touching both the x-axis and the y-axis.What are the coordinates of its center of gravity?
A)(1.0, 1.0) m
B)(0.7, 0.7) m
C)(0.5, 0.5) m
D)Since there is nothing at the center of the hoop, it has no center of gravity.

Accepted Answer

The answer of A hoop of radius 1.0 m is...

Question 8

A uniform bridge span weighs 50.0 * 10³ N and is 40.0 m long.An automobile weighing 15.0 * 10³ N is parked with its center of gravity located 12.0 m from the right pier.What upward support force does the left pier provide? A)29.5 * 10³N B)35.5 * 10³ N C)65.0 * 10³N D)32.5 * 10³ N

Accepted Answer

The answer of A uniform bridge span weighs 50.0 *...

Question 9

Masses are distributed in the x,y-plane as follows: 6.0 kg at (0.0, 0.0) m, 4.0 kg at (2.0, 0.0) m, and 5.0 kg at (2.0, 3.0) m.What is the x-coordinate of the center of gravity of this system of masses?
A)18 m
B)2.0 m
C)1.2 m
D)1.0 m

Accepted Answer

The x-coordinate of the center of gravity is given by the formula:

x̄ = (∑mixi) / (∑mi)

where mi is the mass of each object and xi is its x-coordinate.

Plugging in the values from the problem, we get:

x̄ = [(6.0 kg)(0.0 m) + (4.0 kg)(2.0 m) + (5.0 kg)(2.0 m)] / (6.0 kg + 4.0 kg + 5.0 kg)
x̄ = 1.2 m

Therefore, the x-coordinate of the center of gravity of this system of masses is 1.2 m. Therefore, the answer is C.

Question 10

Masses are distributed in the xy-plane as follows: 10 kg at (2.0, 6.0) m, 4.0 kg at (2.0, 0.0) m, and 6.0 kg at (0.0, 3.0) m.Where would a 20-kg mass need to be positioned so that the center of gravity of the resulting four mass system would be at the origin?
A)(1.4, 3.9) m
B)(3.9, 1.4) m
C)(-1.4, -3.9) m
D)(-3.9, -1.4) m

Accepted Answer

The answer of Masses are distributed in the xy-plane as...

Question 11

The quantity "moment of inertia" (in terms of the fundamental quantities of mass, length, and time) is equivalent to: A)ML²T^-2. B)ML. C)ML². D)ML^-1T^-2.

Accepted Answer

The moment of inertia is defined as I = MR², where M is mass and R is the distance from the axis of rotation. Therefore, the units of moment of inertia are (mass) x (distance)². Using the fundamental units, this can be written as ML². Therefore, choice C is the correct option. The other choices include unnecessary terms or incorrect exponents.

Question 12

An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end.If the scaffold weighs 500 N and the worker stands 1.0 m from one end, what is the tension in the rope nearest the worker?
A)450 N
B)500 N
C)800 N
D)850 N

Accepted Answer

We can start by drawing a diagram and labeling the forces acting on the scaffold. Since the scaffold is in equilibrium (not accelerating and not rotating), the net force and net torque must both be zero.

We can start by summing the forces in the vertical direction (since the scaffold is not accelerating vertically):
Tension closer to worker + Tension closer to other end + Weight of scaffold + Weight of worker = 0

We can simplify this equation by solving for one of the tensions:
Tension closer to worker = -(Tension closer to other end + Weight of scaffold + Weight of worker)

Plugging in the given values, we get:
Tension closer to worker = -(Tension closer to other end + 500 N + 800 N)

Next, we can sum the torques around one of the vertical ropes, say the one closer to the worker. Since the scaffold is in equilibrium, the net torque must be zero:
(Torque due to tension closer to worker) - (Torque due to tension closer to other end) - (Torque due to weight of scaffold) - (Torque due to weight of worker) = 0

We can simplify this equation using the torque formula (torque = force x lever arm):
(Tension closer to worker)(3 m) - (Tension closer to other end)(4 m) - (500 N)(2 m) - (800 N)(1 m) = 0

Plugging in the expression we found for tension closer to worker in the first equation, we get:
-(Tension closer to other end + 500 N + 800 N)(3 m) - (Tension closer to other end)(4 m) - (500 N)(2 m) - (800 N)(1 m) = 0

Now we can solve for tension closer to other end:
Tension closer to other end = 850 N

Finally, we can use the first equation to find tension closer to worker:
Tension closer to worker = -(850 N + 500 N + 800 N) = -2150 N

Since tension is a positive force, we know the actual tension must be positive, so we take the absolute value and get:
Tension closer to worker = 2150 N

However, the question asks for the tension nearest to the worker, which is the tension we solved for as tension closer to other end. Therefore, the answer is:
Answer: D (850 N)

Question 13

A vault is opened by applying a force of 300 N perpendicular to the plane of the door, 0.80 m from the hinges.Find the torque due to this force about an axis through the hinges.
A)120 N*m
B)240 N*m
C)300 N*m
D)360 N*m

Accepted Answer

The answer of A vault is opened by applying a...

Question 14

Tasha has mass 20 kg and wants to use a 4.0-m board of mass 10 kg as a seesaw.Her friends are busy, so Tasha seesaws by herself by putting the support at the system's center of gravity when she sits on one end of the board.How far is she from the support point?
A)2.0 m
B)1.0 m
C)0.67 m
D)0.33 m

Accepted Answer

To balance the seesaw, the sum of the torques on either side of the pivot point must be equal. We can use the equation for torque:
Torque = force x distance from pivot point
Since Tasha is sitting at the end of the board, her weight will act as the force causing the torque. The weight can be calculated as:
Weight = mass x gravity
Weight = 20 kg x 9.8 m/s^2 = 196 N
To balance the seesaw, the torque she creates must be equal to the torque created by the board's weight. The torque created by the board can be calculated as:
Torque = weight of board x distance from pivot point
Torque = 10 kg x 9.8 m/s^2 x distance from pivot point
Setting these two torques equal to each other gives:
force x distance from pivot = weight of board x distance from pivot point
Simplifying gives:
distance from pivot = (weight of board / force) x distance from pivot
Plugging in the values:
distance from pivot = (10 kg x 9.8 m/s^2) / 196 N x 4.0 m
distance from pivot = 0.67 m

Question 15

A uniform, horizontal beam of length 6.0 m and weight 120 N is attached at one end to a wall by a pin connection (so that it may rotate).A cable attached to the wall above the pin supports the opposite end.The cable makes an angle of 60$\degree$with the horizontal.What is the tension in the cable needed to maintain the beam in equilibrium?
A)35 N
B)69 N
C)60 N
D)120 N

Accepted Answer

Let's draw a diagram:

![beam diagram](https://i.imgur.com/qbOtmsy.png)

We can see that there are two forces acting on the beam: 
- its weight, W = 120 N, acting downwards at its center of mass
- the tension in the cable, T, acting upwards at an angle of 60°

Since the beam is in equilibrium, the net force and net torque on the beam must be zero.

Let's first consider the net force: 
- in the horizontal direction, there is no force since the beam is not accelerating in that direction 
- in the vertical direction, the net force is given by Tsin60° - W = 0 
    - solving for T, we get T = W/sin60° = 120 N / sin60° ≈ 138.6 N

Now let's consider the net torque about the pin connection: 
- we can choose any point to calculate torque, but it is often convenient to choose the point where one of the forces acts (in this case, the pin connection) 
- we can take counterclockwise torques as positive 
- the weight of the beam creates a clockwise torque of (W/2) * 6.0 m = 360 Nm 
- the tension in the cable creates a counterclockwise torque of T * 6.0 m * cos60° = 3.0T Nm (since the perpendicular distance from the cable to the pin connection is 6cos60° = 3.0 m) 
- the net torque is given by 3.0T - 360 = 0 
    - solving for T, we get T = 360/3.0 ≈ 120 N

We have two different values for T, but note that the first value (138.6 N) was calculated from the net force, while the second value (120 N) was calculated from the net torque. The correct value for T is the one that satisfies both equations, which is T = 120 N.

Therefore, the answer is B) 69 N.

Question 16

Two children seat themselves on a seesaw.The one on the left has a weight of 400 N while the one on the right weighs 300 N.The fulcrum is at the midpoint of the seesaw.If the child on the left is not at the end but is 1.50 m from the fulcrum and the seesaw is balanced, what is the torque provided by the weight of the child on the right?
A)600 N*m
B)450 N*m
C)-600 N*m
D)-450 N*m

Accepted Answer

Since the seesaw is balanced, the torques on either side of the fulcrum must be equal. Using the formula for torque (torque = force x perpendicular distance from fulcrum), we can calculate that the torque provided by the child on the left is:
torque = 400 N x 1.50 m = 600 Nm
Therefore, the torque provided by the child on the right must also be 600 Nm (in the opposite direction), in order to balance the seesaw. Since the child on the right has a weight of 300 N, the force they are exerting to create this torque is 300 N. Using the same formula for torque, we can solve for the distance from the fulcrum:
torque = force x distance from fulcrum
600 Nm = 300 N x distance from fulcrum
distance from fulcrum = 2.00 m
Since the child on the right is located 2.00 m from the fulcrum, and the seesaw is balanced, the torque provided by their weight is:
torque = 300 N x 2.00 m = 600 Nm
However, since the torque provided by the child on the right is in the opposite direction to that provided by the child on the left, the answer is actually -600 Nm or -600 NF1F1F1S1F1F1F10m. Therefore, the best choice is C.

Question 17

An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall.If the ladder has a mass of 20 kg and it makes an angle of 60$\degree$with the ground, find the force of friction of the ground on the foot of the ladder. A)7.8 * 10² N B)2.0 * 10² N C)50 N D)1.7 * 10² N

Accepted Answer

First, we need to find the force of tension in the ladder. We can do this by balancing the forces in the vertical direction: 
T * cos60° - (80 kg + 20 kg) * g = 0 
T * cos60° = 1000 N 
T = 2000/√3 N

Next, we need to find the force of friction at the foot of the ladder. We can do this by balancing the forces in the horizontal direction: 
Ff + T * sin60° = 0 
Ff = - T * sin60°

Substituting the value of T we found earlier, we get: 
Ff = - 1000 N

Since the force of friction cannot be negative, we take the magnitude of this value: 
Ff = 1000 N

Rounding this to one significant figure, we get the answer: 
Ff = 1.0 x 103 N, which is closest to option D, 1.7 x 1020 N.

Question 18

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well.A crank with a turning radius of 0.25 m is attached to the end of the cylinder.What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g = 9.8 m/s².) A)45 N B)68 N C)90 N D)135 N

Accepted Answer

The answer of A bucket of water with total mass...

Question 19

A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N.The board is supported at each end.The support force at the right end is 3 times the support force at the left end.How far from the right end is the woman standing?
A)4.0 m
B)2.0 m
C)2.7 m
D)1.6 m

Accepted Answer

The answer of A woman who weighs 500 N stands...

Question 20

A rod of length L is pivoted about its left end and has a force F applied perpendicular to the other end.The force F is now removed and another force F' is applied at the midpoint of the rod.If F' is at an angle of 30$\degree$ with respect to the rod, what is its magnitude if the resulting torque is the same as when F was applied?
A)F
B)2F
C)3F
D)4F

Accepted Answer

The answer of A rod of length L is pivoted...

Quiz 8: Rotational Equilibrium and Rotational Dynamics

A bucket filled with water has a mass of 23 kg and is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well.What torque does the weight of water and bucket produce on the cylinder if the cylinder is not permitted to rotate? (g = 9.8 m/s²)

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end.How far would the worker stand from one of the supporting ropes to produce a tension of 550 N in that rope?

A meter stick is supported by a knife-edge at the 50-cm mark.Doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks, respectively.Where should Doug hang a third mass of 0.30 kg to keep the stick balanced?

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall.The coefficient of static friction between ladder and floor is 0.40.What minimum angle can the ladder make with the floor before it slips?

A hoop of radius 1.0 m is placed in the first quadrant of an xy-coordinate system with its rim touching both the x-axis and the y-axis.What are the coordinates of its center of gravity?

A uniform bridge span weighs 50.0 * 10³ N and is 40.0 m long.An automobile weighing 15.0 * 10³ N is parked with its center of gravity located 12.0 m from the right pier.What upward support force does the left pier provide?

Masses are distributed in the x,y-plane as follows: 6.0 kg at (0.0, 0.0) m, 4.0 kg at (2.0, 0.0) m, and 5.0 kg at (2.0, 3.0) m.What is the x-coordinate of the center of gravity of this system of masses?

Masses are distributed in the xy-plane as follows: 10 kg at (2.0, 6.0) m, 4.0 kg at (2.0, 0.0) m, and 6.0 kg at (0.0, 3.0) m.Where would a 20-kg mass need to be positioned so that the center of gravity of the resulting four mass system would be at the origin?

The quantity "moment of inertia" (in terms of the fundamental quantities of mass, length, and time) is equivalent to:

An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end.If the scaffold weighs 500 N and the worker stands 1.0 m from one end, what is the tension in the rope nearest the worker?

A vault is opened by applying a force of 300 N perpendicular to the plane of the door, 0.80 m from the hinges.Find the torque due to this force about an axis through the hinges.

Tasha has mass 20 kg and wants to use a 4.0-m board of mass 10 kg as a seesaw.Her friends are busy, so Tasha seesaws by herself by putting the support at the system's center of gravity when she sits on one end of the board.How far is she from the support point?

An 80-kg man is one fourth of the way up a 10-m ladder that is resting against a smooth, frictionless wall.If the ladder has a mass of 20 kg and it makes an angle of 60 $\degree$ with the ground, find the force of friction of the ground on the foot of the ladder.

A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N.The board is supported at each end.The support force at the right end is 3 times the support force at the left end.How far from the right end is the woman standing?