A block with mass mi hangs from a rope that is extended over an ideal pulley and

Question

A block with mass mi hangs from a rope that is extended over an ideal pulley and attached to a second block with mass m_2 that sits on a ledge slanted at an angle of 20 degree (see figure below). Suppose the system of blocks is initially held motionless and, when released, begins to accelerate. If m_1 = 6.50 kg, m_2 = 3.50 kg, and the magnitude of the acceleration of the blocks is 0.170 m/s^2, find the magnitude of the kinetic frictional force between the second block and the ledge. What is the value of the coefficient of kinetic friction between the block and the ledge?

Explanation / Answer

let tension in the string be T.

part a:

as m1>m2, lets assume that m1 is going down and m2 is going up.

then friction force will be acting along the incline and in downward direction

normal force on the block m2=m2*g*cos(20)=32.23 N

component of weight of m2 along the incline=m2*g*sin(20)=11.7313 N

let friction force be F.

applying force balance equation for m1:

m1*g-T=m1*0.17

==>T=62.595 N

then applying force balance equation for m2:

T-11.7313-F=3.5*0.17

==>F=T-11.7313-3.5*0.17=50.2687 N

hence friction force magnitude is 50.2687 N

part b:

coefficient of kinetic friction=friction force/normal force=50.2687/32.23=1.5596

note:

coefficient of friction can never be higher than 1. so please check the data in question if there is any mistake in printing .

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.