Question

A wooden block with mass 0.37 kg rests on a horizontal table, connected to a string that hangs vertically over a friction-less pulley on the table's edge. From the other end of the string hangs a 0.12 kg mass.

What minimum coefficient of static friction μ s between the block and table will keep the system at rest?

Find the block's acceleration if μ k =0.20.

Answer 1

Mass of block on the table = m1 = 0.37 kg

Mass of block hanging to the string = m2 = 0.12 kg

Coefficient of friction = µ

f = f_{s, max} = µ N

Acceleration of Block m1 = a = 0

Hence,   T - f_{s, max} = 0

T = µ N

Acceleration of Block m2 = 0

Hence,   T = m₂ g = 0.12 * 9.8 = 1.176 N

Therefore µ =   T / N   = T / (m1 x g)   = 1.176 / (0.37 x 9.8) = 0.324

The block's acceleration if μ k =0.20

The tension on the string due to block m1 = Tension on the string due to block m2

m1 ( a + μ k g ) = m2 ( g – a)

a = (m2 g - m1 μ k g) / (m1 + m2)

    = 0.91 m / s²