The masses are m A =15 kg and m B =30 kg. The coefficients of friction for all s
ID: 1698823 • Letter: T
Question
The masses are mA=15 kg and mB=30 kg. The coefficients of friction for all surfaces are (static) s=0.4 and (kinetic) k=0.35.
If P=400 N, what are the accelerations (m/s2) of A and B?
The surface is completely horizontal, mA is on top of mB and the force (P) acts on mB to the right. (I tried to draw a diagram, but it won't let me post it)
Explanation / Answer
from free body diagram we can write fs,max = mA * a1 Fn1 = mA * g For lower block Fmax - fs,max = m2 * a2 From above relations we can write a1 = us * g = 0.4 * 9.8 = 3.92 m/s^2 and Fmax = (mA + mB) * us * g = (15 + 30) * 0.4 * 9.8 = 176.4 N b) from free body diagram we can write fk = mA * a1 ==> a1 = uk * g = 0.35 * 9.8 = 3.43 m/s^2 Fn1 = mA * g For lower block Fmax - mA*uk * g = m2 * a2 ==> a2 = (Fmax - uk mA*g) / mB = [176.4 - 51.45] / 35 = 3.57 m/s^2
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