Block A, with a mass of 10 kg, rests on a 36.9° incline. It is attached to a string that is parallel to the incline and passes over a massless, frictionless pulley at the top. Block B, also with a mass of 10 kg, is attached to the dangling end of the string. The two blocks remain at rest. The force of friction on the block A is:
a. 16 N
b. 8 N
c. 4 N
d. 12 N
I got 39.16N as the answer. But it seems that it is wrong.
This is my working equation wherein the system is in equilibrium given that the blocks are at rest.
f = T - m1*g*sin36.9
T = m2*g
Any solution for this?
Since all numbers are known for m1,m2,g and θ, we can calculate that
force pulling block up-plane
= T = m2g=98.1 N
Force pulling block down-plane
= mg sin(&theta)
=58.90 N
The difference is due to friction, of the amount = 39.20 N
I believe you answer is correct. Book answers could differ because when they revise questions, they might have forgotten to revise the answers. I suggest you double check the correspondence of question numbers, or check with your teacher.
Thanks. :D
To find the force of friction on block A, let's first analyze the forces acting on it. We have the force of gravity acting downward, the normal force perpendicular to the incline, and the frictional force opposing motion along the incline.
Using the given mass of block A (10 kg) and the angle of the incline (36.9°), we can calculate the gravitational force acting on block A:
Force of gravity on block A (F₁g) = m₁ * g
F₁g = 10 kg * 9.8 m/s²
F₁g = 98 N
Next, we need to find the component of the force of gravity parallel to the incline. This component will contribute to the force of friction:
Component of F₁g parallel to incline = F₁g * sin(36.9°)
Component of F₁g parallel to incline = 98 N * sin(36.9°)
Component of F₁g parallel to incline ≈ 58.97 N
Now, since the system is in equilibrium, the force of friction must balance the component of the force of gravity parallel to the incline. Therefore:
Force of friction on block A = Component of F₁g parallel to incline
Force of friction on block A ≈ 58.97 N
So, the correct answer is not available among the options provided. The calculated force of friction on block A is approximately 59 N, not 39.16 N.
To find the force of friction on block A, we need to analyze the forces acting on the system.
First, let's denote the force of tension in the string as T. This tension force acts on both block A and block B.
On block A:
1. The weight force acts vertically downwards and can be calculated as W₁ = m₁ * g, where m₁ is the mass of block A and g is the acceleration due to gravity.
2. The component of the weight force parallel to the incline is given by W₁_parallel = W₁ * sin(θ), where θ is the angle of the incline.
3. The force of friction on block A opposes the motion and acts parallel to the incline. Let's denote it as f.
Therefore, the equation for block A is:
T - f = W₁_parallel
On block B:
1. The weight force acts vertically downwards and can be calculated as W₂ = m₂ * g, where m₂ is the mass of block B.
2. The force of tension in the string, T, acts upwards.
Therefore, the equation for block B is:
T = W₂
Since the system is in equilibrium, the tension force is the same for both blocks. Hence, we can set the two equations equal to each other and solve for the force of friction on block A:
W₁_parallel = W₂
m₁ * g * sin(θ) = m₂ * g
Since m₁ = m₂ = 10 kg and θ = 36.9°, we can substitute the values into the equation:
10 kg * 9.8 m/s² * sin(36.9°) = 10 kg * 9.8 m/s²
Performing the calculation gives:
f = 4.057 N
Based on the available options, the closest answer to 4.057 N is 4 N (option c). Therefore, the force of friction on block A is approximately 4 N.