Answer:
a. 94.54 N
b. 0.356 m/s^2
Explanation:
Given:-
- The mass of the inclined block, M = 100 kg
- The mass of the vertically hanging block, m = 10 kg
- The angle of inclination, θ = 20°
- The coefficient of friction of inclined surface, u = 0.3
Find:-
a) The magnitude of tension in the cable
b) The acceleration of the system
Solution:-
- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.
- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.
- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:
... Eq 1
Where,
a: The acceleration of the system
- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.
- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.
- We will employ the static equilibrium of the inclined block in the normal direction and we have:
![N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )](https://tex.z-dn.net/?f=N%20-%20M%2Ag%2Acos%20%28%20Q%20%29%3D%200%5C%5C%5C%5CN%20%3D%20M%2Ag%2Acos%20%28%20Q%20%29)
- The frictional force ( F ) is proportional to the contact force ( N ) as follows:
![F = u*N\\\\F = u*M*g *cos ( Q )](https://tex.z-dn.net/?f=F%20%3D%20u%2AN%5C%5C%5C%5CF%20%3D%20u%2AM%2Ag%20%2Acos%20%28%20Q%20%29)
- Now we will apply the Newton's second law of motion parallel to the plane as follows:
.. Eq2
- Add the two equation, Eq 1 and Eq 2:
![M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}](https://tex.z-dn.net/?f=M%2Ag%2Asin%20%28%20Q%20%29%20-%20u%2AM%2Ag%2Acos%20%28%20Q%20%29%20-%20m%2Ag%20%3D%20a%2A%20%28%20M%20%2B%20m%20%29%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7BM%2Ag%2Asin%20%28%20Q%20%29%20-%20u%2AM%2Ag%2Acos%20%28%20Q%20%29%20-%20m%2Ag%7D%7BM%20%2B%20m%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B100%2A9.81%2Asin%20%28%2020%20%29%20-%200.3%2A100%2A9.81%2Acos%20%28%2020%20%29%20-%2010%2A9.81%7D%7B100%20%2B%2010%7D%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B-39.12977%7D%7B110%7D%20%3D%20-0.35572%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).
- Using equation 1, we determine the tension ( T ) in the cable as follows:
![T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N](https://tex.z-dn.net/?f=T%20%3D%20m%2A%20%28%20a%20%2B%20g%20%29%5C%5C%5C%5CT%20%3D%2010%2A%28%20-0.35572%20%2B%209.81%20%29%5C%5C%5C%5CT%20%3D%2094.54%20N)