Question

Consider the two-body system at the right. A 22.7-N block is placed upon an inclined plane which is inclined at a 17.2 degree angle. The block is attached by a string to a 34.5-N block which is suspended over a pulley. The coefficient of friction is 0.219. Determine the acceleration of the block and the tension in the stringAnswers:

Answer 1

Answer:

Given that

For A weight Wt= 22.7 N

m₁ = 22.7/10 = 2.27 kg

Force alone inclined plane

Wt₁ = m₁ g sin θ

Wt₁  =22.7 sin 17.2°

Wt₁ = 6.7 N

For B weight Wt₂= 34.5-N

m₂ = 3.45 kg

coefficient of friction ,μ= 0.219

θ = 17.2 degree

The friction force on the block A

Fr= μ m₁ g cos θ

Fr= 0.216 x 22.7 x cos 17.2°

Fr= 4.68 N

Lets take acceleration of system is a m/s²

Tension = T

From Newtons law

Wt₂ - Wt₁ - Fr = (m₁ +m₂) a

34.5 -  6.7 - 4.68 = (2.27 + 3.45 ) a

a= 4.05 m/s²

Block B

Wt₂ - T = m₂ a

T = 34.5 - 4.05 x 3.45

T= 20.52 N