A block of mass m 1 = 2.00 kg and a block of mass m 2 = 6.30 kg are connected by

Question

A block of mass m1 = 2.00 kg and a block of mass m2 = 6.30 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0°. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley.

(a) Determine the acceleration of the two blocks.
m/s2

(b) Determine the tensions in the string on both sides of the pulley.

Explanation / Answer

given,

ma = 2 kg

m2 = 6.3 kg

R = 2.5 kg

M = 10 kg

theta = 30 degree

coefficient of kinetic friction = 0.36

equation of motion of m1

T1 - mu * m1 * g = m1 * a

T1 - 0.36 * 2 * 9.8 = 2 * a

equation of motion of m2

m2 * g * sin(theta) - mu * m2 * g * cos(theta) - T2 = m2 * a

6.3 * 9.8 * sin(30) - 0.36 * 6.3 * 9.8 * cos(30) - T2 = 6.3 * a

equation of motion of pulley

T2 * R - T1 * R = I * a / R

T2 * R - T1 * R = 0.5 * M * R^2 * a / R

T2 * 0.25 - T1 * 0.25 = 0.5 * 10 * 0.25 * a

on solving we'll get

acceleration a = 0.343261 m/s^2

tension left of pulley T1 = 7.74252 N

tension right of pulley T2 = 9.45883 N

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