I asked this question earlier, but the work provided didn\'t really make sense t

Question

I asked this question earlier, but the work provided didn't really make sense to me or explain how the problem was solved. Too many steps were skipped. Please be as thourough as possible.

Box 1 is connected to box 2 with a perfect rope over a massless and frictionless pulley. Find the acceleration of box 1 and the tension in the rope connecting box 1 and box 2. The coefficient of kinetic friction between box 1 and the surface is 0.06 and the coefficient of static friction between box 1 and the surface is 0.1. The mass of box 1 is 5kg and the mass of box 2 is 10kg.

Explanation / Answer

coefficient of static friction between box and floor, ks = 0.1
coeffiient of kinetic friciton, k = 0.06

now, tension in the rope = T
acceleration of the system = a ( downwards)

so, for the hanging mass
m2*g - T = m2*a
for the sliding mass
T - f = m1*a
here f is force of friction
f = k*m1*g
hence
T - k*m1*g = m1*a
hence

m2g - k*m1*g = (m1 + m2)a
a = (m2 - k*m1)g/(m1 + m2)
and T = m1*(m2 - km1)g/(m1 + m2) + k*m1*g

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