Question 1

A 75.0 kg ladder that is 3.00 m long is placed against a wall at an angle theta. The center of gravity of the ladder is at a point 1.2 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.400. There is no friction between the wall and the ladder. What is the minimum angle the ladder must make with the horizontal for the ladder not to slip and fall? 
A) 35 degrees
B) 45 degrees
C) 53 degrees
D) 60 degrees
E) 65 degrees

Accepted Answer

The answer of A 75.0 kg ladder that is 3.00...

Question 2

A 75.0 kg ladder that is 3.00 m long is placed against a wall at an angle theta. The center of gravity of the ladder is at a point 1.20 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.800. There is no friction between the wall and the ladder. What is the minimum angle the ladder must make with the horizontal for the ladder not to slip and fall? 
A) 26.57 degrees
B) 30.34 degrees
C) 36.35 degrees
D) 40.55 degrees
E) 46.52 degrees

Accepted Answer

The answer of A 75.0 kg ladder that is 3.00...

Question 3

A 75.0 kg ladder that is 3.00 m long is placed against a wall at an angle theta. The center of gravity of the ladder is at a point 1.20 m from the base of the ladder. The coefficient of static friction at the base of the ladder is 0.800. There is no friction between the wall and the ladder. What is the vertical force of the ground on the ladder? 
A) 625 N
B) 640 N
C) 735 N
D) 832 N
E) 900 N

Accepted Answer

First, we need to find the force of gravity acting on the ladder. Using the formula Fg = mg, where m is the mass of the ladder and g is the acceleration due to gravity, we get:

Fg = (75.0 kg)(9.81 m/s^2) = 736 N

Next, we need to find the force acting to the left on the ladder due to friction. This force will be equal to the force of friction, which we can find using the formula Ff = μN, where μ is the coefficient of static friction and N is the normal force acting on the ladder. The normal force can be found using the fact that the ladder is in static equilibrium, so the sum of the forces in the y-direction must be zero. Thus:

N + Fgy = Ffy
N = Ffy - Fgy

where Fgy is the force of gravity acting in the y-direction and Ffy is the force of the ground on the ladder in the y-direction. Since the ladder is not moving in the y-direction, Ffy = N. The force of gravity in the y-direction can be found using trigonometry:

Fgy = (mg/L) * x = [(75.0 kg)(9.81 m/s^2)/3.00 m] * 1.20 m = 294.3 N

where L is the length of the ladder and x is the distance from the base of the ladder to the center of gravity. Plugging in the numbers, we get:

N = Ffy = Fgy - μN
N = 294.3 N - 0.800N
N = 612.5 N

Finally, we can find the force of the ground on the ladder in the x-direction using the fact that the ladder is in static equilibrium, so the sum of the forces in the x-direction must be zero. Thus:

Fx = Ff = μN = 0.800(612.5 N) = 490 N

The vertical force of the ground on the ladder is then given by the Pythagorean theorem:

F = sqrt(Fx^2 + Ffy^2) = sqrt((490 N)^2 + (612.5 N)^2) = 735 N

Therefore, the answer is (C).

Question 4

A 1.500 m long uniform beam of mass 30.00 kg is supported by a wire as shown in the figure. The beam makes an angle of 10.00 degrees with the horizontal and the wire makes an angle of 30.00 degrees with the beam. A 50.00 kg mass, m, is attached to the end of the beam. What is the tension in the wire? 
A) 2,034 N
B) 1,855 N
C) 1,435 N
D) 1,255 N
E) 1,035 N

Accepted Answer

The answer of A 1.500 m long uniform beam of...

Question 5

A 6.00 kg mass is located at (1.0 m, −2.00 m, 3.00 m), a 5.00 kg mass is located at (1.0 m, 3.00 m, −2.00 m), and a 4.00 kg mass is located at (−1.0 m, −2.00 m, 2.00 m). The center of gravity of the system of masses is
A) (5/15 m, −1/15 m, 17/15 m).
B) (5/15 m, −17/15 m, 5/15 m).
C) (12/15 m, −5/15 m, 16/15 m).
D) (7/15 m, −5/15 m, 16/15 m).
E) (16/15 m, −1/15 m, 17/15 m).

Accepted Answer

The center of gravity is calculated by taking the weighted average of the positions of the masses. The x, y, and z coordinates are calculated separately using the formula: (Σmi*xi)/Σmi, (Σmi*yi)/Σmi, (Σmi*zi)/Σmi. For this system, the center of gravity is (7/15 m, −5/15 m, 16/15 m).

Question 6

A 10 kg object has a moment of inertia of 1.25 kg m². If a torque of 2.5 N•m is applied to the object, the angular acceleration is A) 10 rad/s². B) 8.0 rad/s². C) 6.0 rad/s². D) 4.0 rad/s². E) 2.0 rad/s².

Accepted Answer

The answer of A 10 kg object has a moment...

Question 7

A 1.500 m long uniform beam of mass 30.00 kg is supported by a wire as shown in the figure. The beam makes an angle of 10.00 degrees with the horizontal and the wire makes an angle of 30.00 degrees with the beam. A 50.00 kg mass, m, is attached to the end of the beam. What are the vertical and horizontal components of the force of the wall on the beam at the hinge? 
A) H = 1,458 N, V = 454.0 N
B) H = 1,300 N, V = 403.4 N
C) H = 1,179 N, V = 354.9 N
D) H = 979 N, V = 324.5 N
E) H = 750 N, V = 297.3 N

Accepted Answer

The answer of A 1.500 m long uniform beam of...

Question 8

A 5.00 kg mass is located at (1.0 m, 0.00 m, 3.00 m), a 2.00 kg mass is located at (0.00 m, 3.00 m, −2.00 m), and a 3.00 kg mass is located at (−1.0 m , −2.00 m , 0.00 m). The center of gravity of the system of masses is
A) (1/10 m, 10/10 m, 1/10 m).
B) (2/10 m, 0 m, 11/10 m).
C) (3/10 m, 2/10 m, 10/10 m).
D) (10/10 m, 2/10 m, 3/10 m).
E) (2/10 m, 10/10 m, 0 m).

Accepted Answer

The center of gravity (CG) of a system of masses can be found using the formula $\text{CG} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$, where $m_i$ is the mass of the $i$th object and $\vec{r}_i$ is the position vector of the $i$th object. For the x-coordinate of the CG, $\text{CG}_x = \frac{(5.00\,kg)(1.0\,m) + (2.00\,kg)(0.00\,m) + (3.00\,kg)(-1.0\,m)}{5.00\,kg + 2.00\,kg + 3.00\,kg} = \frac{2\,m}{10\,kg} = \frac{2}{10}\,m$. For the y-coordinate, $\text{CG}_y = \frac{(5.00\,kg)(0.00\,m) + (2.00\,kg)(3.00\,m) + (3.00\,kg)(-2.00\,m)}{10\,kg} = \frac{0\,m}{10\,kg} = 0\,m$. For the z-coordinate, $\text{CG}_z = \frac{(5.00\,kg)(3.00\,m) + (2.00\,kg)(-2.00\,m) + (3.00\,kg)(0.00\,m)}{10\,kg} = \frac{11\,m}{10\,kg} = \frac{11}{10}\,m$. Therefore, the center of gravity is at $(\frac{2}{10}\,m, 0\,m, \frac{11}{10}\,m)$.

Question 9

Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weighs 800 N. Wayne's center of gravity is 75.0 cm from Chris. Chris and Jamie are at the ends of the stretcher. The upward force that Chris is exerting to support the stretcher, with Wayne on it, is
A) 250 N.
B) 350 N.
C) 400 N.
D) 550 N.
E) 650 N.

Accepted Answer

The answer of Chris and Jamie are carrying Wayne on...

Question 10

An 8.00 kg object has a moment of inertia of 1.50 kg m². If a torque of 2.00 N•m is applied to the object, the angular acceleration is A) 0.750 rad/s². B) 1.00 rad/s². C) 1.33 rad/s². D) 2.01 rad/s². E) 2.67 rad/s².

Accepted Answer

We can use the equation τ = Iα (torque equals moment of inertia times angular acceleration) to solve for the angular acceleration. Rearranging the equation, we get: α = τ/I Plugging in the values given in the problem, we get: α = 2.00 N•m / 1.50 kg m² α = 1.33 rad/s² Therefore, the angular acceleration is 1.33 rad/s². The correct answer is choice C).

Question 11

Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weighs 800 N. Wayne's center of gravity is 75.0 cm from Chris. Chris and Jamie are at the ends of the stretcher. The upward force that Jamie is exerting to support the stretcher, with Wayne on it, is
A) 250 N.
B) 300 N.
C) 350 N.
D) 400 N.
E) 550 N.

Accepted Answer

First, we need to find the distance between Jamie and Wayne's center of gravity. This can be found by subtracting the distance between Chris and Wayne's center of gravity (which is given to be 75 cm or 0.75 m) from the total length of the stretcher (2.00 m):

distance between Jamie and Wayne's center of gravity = 2.00 m - 0.75 m = 1.25 m

Next, we need to determine the force that each person is exerting on the stretcher. Since the stretcher is not accelerating vertically (i.e. it is not going up or down), the sum of the forces in the vertical direction must be zero:

F_chris + F_jamie - weight_of_stretcher - weight_of_Wayne = 0

F_chris + F_jamie - 900 N = 0

F_chris + F_jamie = 900 N

Since Chris and Jamie are at opposite ends of the stretcher and are exerting equal and opposite forces on the stretcher (since they are not accelerating horizontally), we can assume that they are each exerting half of the total force:

F_chris = F_jamie = 450 N

Finally, we need to determine the upward force that Jamie is exerting to support the stretcher with Wayne on it. Since Wayne's weight is not distributed evenly along the stretcher (his center of gravity is not at the center of the stretcher), Jamie will need to exert more upward force than Chris in order to keep the stretcher level. We can use the principle of moments to determine this force:

moment_of_F_chris_about_Jamie = F_chris x distance_between_Jamie_and_Chris

moment_of_Wayne_about_Jamie = weight_of_Wayne x distance_between_Jamie_and_Wayne's_center_of_gravity

Since the stretcher is not rotating, the sum of the moments about Jamie must be zero:

moment_of_F_chris_about_Jamie - moment_of_Wayne_about_Jamie = 0

F_chris x distance_between_Jamie_and_Chris - weight_of_Wayne x distance_between_Jamie_and_Wayne's_center_of_gravity = 0

450 N x 0.75 m - 800 N x 1.25 m = 0

337.5 N - 1000 N = 0

-662.5 N = 0

This is obviously not possible, so we need to reverse the sign of the moment of Wayne about Jamie:

moment_of_F_chris_about_Jamie - (-moment_of_Wayne_about_Jamie) = 0

F_chris x distance_between_Jamie_and_Chris + weight_of_Wayne x distance_between_Jamie_and_Wayne's_center_of_gravity = 0

450 N x 0.75 m + 800 N x 1.25 m = 0

337.5 N + 1000 N = 0

662.5 N = 0

Now that we have the moment equation set up correctly, we can solve for the upward force that Jamie is exerting:

F_jamie = (weight_of_stretcher + weight_of_Wayne - F_chris) / 2

F_jamie = (100 N + 800 N - 450 N) / 2

F_jamie = 350 N

Therefore, the upward force that Jamie is exerting to support the stretcher with Wayne on it is 350 N, which corresponds to choice (C).

Question 12

A 5.00 kg object has a moment of inertia of 1.20 kg m². What torque is needed to give the object an angular acceleration of 2.0 rad/s²? A) 2.4 N·m B) 2.6 N·m C) 2.8 N·m D) 3.0 N·m E) 3.2 N·m

Accepted Answer

The torque required can be calculated using the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Substituting the given values, τ = 1.20 kg·m^2 * 2.0 rad/s^2 = 2.4 N·m.

Question 13

A 2.000 m long horizontal uniform beam of mass 20.00 kg is supported by a wire as shown in the figure. The wire makes an angle of 20.00 degrees with the beam. Attached to the beam 1.400 m from the wall is a ball with a mass of 40.00 kg. What are the vertical and horizontal components of the force of the wall on the beam at the hinge? 
A) V = 175.6 N, H = 2,023 N
B) V = 186.6 N, H = 1,805 N
C) V = 195.4 N, H = 1,750 N
D) V = 200.6 N, H = 1,323 N
E) V = 215.6 N, H = 1,023 N

Accepted Answer

The answer of A 2.000 m long horizontal uniform beam...

Question 14

Jim and Mary are carrying Bob on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 80 N. Bob weighs 600 N. Bob's center of gravity is 80 cm from Mary. Jim and Mary are at the ends of the stretcher. The upward force that Mary is exerting to support the stretcher, with Bob on it, is
A) 550 N.
B) 400 N.
C) 300 N.
D) 280 N.
E) 200 N.

Accepted Answer

The answer of Jim and Mary are carrying Bob on...

Question 15

A torque of 2.00 N•m is applied to a 10.0 kg object to give it an angular acceleration. If the angular acceleration is 1.75 rad/s², then the moment of inertia is A) 1.95 kg m². B) 1.05 kg m². C) 1.14 kg m². D) 1.20 kg m². E) 1.35 kg m².

Accepted Answer

The answer of A torque of 2.00 N•m is applied...

Question 16

A 2.00 m long horizontal uniform beam of mass 20.0 kg is supported by a wire as shown in the figure. The wire makes an angle of 20.0 degrees with the beam. Attached to the beam 1.40 m from the wall is a ball with a mass of 40.0 kg. What is the tension in the string? 
A) 1,000 N
B) 1,090 N
C) 2,100 N
D) 2,250 N
E) 2,680 N

Accepted Answer

The answer of A 2.00 m long horizontal uniform beam...

Question 17

A 10 kg solid cylinder with a 50.0 cm radius has a moment of inertia of 1/2 MR². If a torque of 2.0 N•m is applied to the object, the angular acceleration is A) 1.0 rad/s². B) 1.6 rad/s². C) 1.8 rad/s². D) 2.1 rad/s². E) 2.3 rad/s².

Accepted Answer

The formula for torque is given by 
τ = Iα
where τ is the torque, I is the moment of inertia and α is the angular acceleration.
Rearranging the formula gives
α = τ / I
Substituting the given values, we get
α = 2.0 N·m / (1/2 * 10 kg * 0.5 m^2)
α = 1.6 rad/s^2
Therefore, the angular acceleration is 1.6 rad/s^2 or choice B.

Question 18

An 8.0 kg object has a moment of inertia of 1.00 kg m². What torque is needed to give the object an angular acceleration of 1.5 rad/s²? A) 3.0 N·m B) 2.5 N·m C) 2.0 N·m D) 1.5 N·m E) 1.0 N·m

Accepted Answer

The answer of An 8.0 kg object has a moment...

Question 19

A 10 kg sphere with a 25.0 cm radius has a moment of inertia of 2/5 MR². If a torque of 2.0 N•m is applied to the object, the angular acceleration is A) 1.0 rad/s². B) 2.0 rad/s². C) 4.0 rad/s². D) 6.0 rad/s². E) 8.0 rad/s².

Accepted Answer

The answer of A 10 kg sphere with a 25.0...

Question 20

Jim and Mary are carrying Bob on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 80 N. Bob weighs 600 N. Bob's center of gravity is 80 cm from Mary. Jim and Mary are at the ends of the stretcher. The upward force that Jim is exerting to support the stretcher, with Bob on it, is
A) 280 N.
B) 320 N.
C) 380 N.
D) 400 N.
E) 520 N.

Accepted Answer

To find the force that Jim is exerting, we can use the principle of moments (torque) about Mary's end of the stretcher. Let $ F_J $ be the force exerted by Jim. The total weight of Bob and the stretcher is $ 600\,N + 80\,N = 680\,N $. The center of gravity of the system can be found by considering the moments about Mary's end. The stretcher's weight acts at its center, 1.00 m from Mary, and Bob's weight acts 0.80 m from Mary.The moment due to the stretcher's weight is $ 80\,N \times 1.00\,m = 80\,Nm $, and the moment due to Bob's weight is $ 600\,N \times 0.80\,m = 480\,Nm $. The total moment about Mary's end is $ 80\,Nm + 480\,Nm = 560\,Nm $.For equilibrium, the moment about Mary due to Jim's force must equal the total moment due to the weights, so:$$ F_J \times 2.00\,m = 560\,Nm $$$$ F_J = \frac{560\,Nm}{2.00\,m} = 280\,N $$Therefore, Jim is exerting a force of 280 N.

Quiz 8: Torque and Angular Momentum

A 6.00 kg mass is located at (1.0 m, −2.00 m, 3.00 m) , a 5.00 kg mass is located at (1.0 m, 3.00 m, −2.00 m) , and a 4.00 kg mass is located at (−1.0 m, −2.00 m, 2.00 m) . The center of gravity of the system of masses is

A 10 kg object has a moment of inertia of 1.25 kg m². If a torque of 2.5 N•m is applied to the object, the angular acceleration is

A 5.00 kg mass is located at (1.0 m, 0.00 m, 3.00 m) , a 2.00 kg mass is located at (0.00 m, 3.00 m, −2.00 m) , and a 3.00 kg mass is located at (−1.0 m , −2.00 m , 0.00 m) . The center of gravity of the system of masses is

An 8.00 kg object has a moment of inertia of 1.50 kg m². If a torque of 2.00 N•m is applied to the object, the angular acceleration is

A 5.00 kg object has a moment of inertia of 1.20 kg m². What torque is needed to give the object an angular acceleration of 2.0 rad/s²?

A torque of 2.00 N•m is applied to a 10.0 kg object to give it an angular acceleration. If the angular acceleration is 1.75 rad/s², then the moment of inertia is

A 2.00 m long horizontal uniform beam of mass 20.0 kg is supported by a wire as shown in the figure. The wire makes an angle of 20.0 degrees with the beam. Attached to the beam 1.40 m from the wall is a ball with a mass of 40.0 kg. What is the tension in the string?

A 10 kg solid cylinder with a 50.0 cm radius has a moment of inertia of 1/2 MR². If a torque of 2.0 N•m is applied to the object, the angular acceleration is

An 8.0 kg object has a moment of inertia of 1.00 kg m². What torque is needed to give the object an angular acceleration of 1.5 rad/s²?

A 10 kg sphere with a 25.0 cm radius has a moment of inertia of 2/5 MR². If a torque of 2.0 N•m is applied to the object, the angular acceleration is

Quiz 8: Torque and Angular Momentum

A 6.00 kg mass is located at (1.0 m, −2.00 m, 3.00 m) , a 5.00 kg mass is located at (1.0 m, 3.00 m, −2.00 m) , and a 4.00 kg mass is located at (−1.0 m, −2.00 m, 2.00 m) . The center of gravity of the system of masses is

A 10 kg object has a moment of inertia of 1.25 kg m2. If a torque of 2.5 N•m is applied to the object, the angular acceleration is

A 5.00 kg mass is located at (1.0 m, 0.00 m, 3.00 m) , a 2.00 kg mass is located at (0.00 m, 3.00 m, −2.00 m) , and a 3.00 kg mass is located at (−1.0 m , −2.00 m , 0.00 m) . The center of gravity of the system of masses is

An 8.00 kg object has a moment of inertia of 1.50 kg m2. If a torque of 2.00 N•m is applied to the object, the angular acceleration is

A 5.00 kg object has a moment of inertia of 1.20 kg m2. What torque is needed to give the object an angular acceleration of 2.0 rad/s2?

A torque of 2.00 N•m is applied to a 10.0 kg object to give it an angular acceleration. If the angular acceleration is 1.75 rad/s2, then the moment of inertia is