Question

a 20 kg crate sitting on the floor is attached to a rope that pulls 37° above the horizontal the coefficient of friction between the crate and the floor is 0.50

a. construct a free body diagram for the crate
b. determine the minimum rope tension that will cause the crate to start sliding

Answer 1

(1) \[\sum F _{x}= T _{x} - f = m \times a \]
(2) \[\sum F _{x}= T _{y} + N - P = 0\].

step 1. You know that Tx is Tx=(T)x(cos$) and that $= 37° so Tx=(T)x(0.80N)

step 2. You know that Ty is Ty=(T)x(sin$) and that $= 37° so Ty=(T)x(0.60N)

step 3. If f=(coefficient of friction)xN = 0.5xN

step 4. and using the ecuation (2) you find N you get N= P - Ty = mxg - Ty = (20)x(9,8) - (T)x(0.60N) = 196N - (T)x(0.60N) =N

step 5. So now you sub N in the formula that is in step 3. f=(coefficient of friction)xN=0.5xN that is f=98N - (T)x(0.3)

step 6. Now you use the forula (1), subing f=98N - (T)x(0.3) in the ecuation. You get

Tx - f = (T)x(0.80) - [98N - (T)x(0.3)] =mxa
(T)x(0.80) - 98N + (T)x(0.3) =mxa
(T)x(1.1) - 98N = (20kg)x(9,8m/s^2)
(T)x(1.1) - 98N = 196N
(T)x(1.1N) =98N

Then
T= 89.1N