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Fundamentals Of Physics
Found in: Page 140

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Short Answer

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.50 , what minimum force magnitude is required from the rope to start the crate moving? (b) If μk=0.35 , what is the magnitude of the initial acceleration of the crate?

(a) The minimum force magnitude is required from the rope to start the crate moving is 304 N.

(b) The magnitude of the initial acceleration of the crate is 1.3 m/s2.

See the step by step solution

Step by Step Solution

Step 1: Given

Mass of the block, m=68 kg

Coefficient of static friction, μa=0.50

Coefficient of kinetic friction, μk=0.35

Angle, θ=15°

Step 2: Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

Fnet= ma

where, F is the net force, m is mass and a is an acceleration.

Step 3: Determining the free body diagram of the block

Free body diagram of the block:

Step 4: (a) Determining theminimum force required from the rope to start the crate moving

By using Newton’s 2nd law along y direction, block is not moving along y.So, the acceleration is 0, it gives,

Fx=0N-mg+F sin(15)=0N=mg-F sin(15)Similarly, along x direction,Fx=0since block is at rest, a=0Fcos(15)-fs=0fs=Fcos(15)The frictional force is,fs=μsN=μs(mg-Fsin(15))Equating equation (i) and (ii),Fcos(15)=μs(mg-Fsin(15))Fcos(15)+μsFsin(15)=μsmgF=μs mgcos15+μssin15 =0.50689.81cos15+0.50sin15 =304 N

Therefore, the minimum force magnitude is required from the rope to start the crate moving is 304N.

Step 5: (b) Determining themagnitude of the initial acceleration of the crate

By using Newton’s 2nd law along y direction,block is not moving along y.So, the acceleration is 0, it gives,

Fy=0N=mg+F sin15+0N=mg-F

The kinetic frictional force is,

fk=μkN =μkmg-Fsin15

According to the Newton’s 2nd law along horizontal direction i.e. x is as the block movingalong x.Let, the acceleration of the block is a,

Fx =maxFcos15-fx=maa=Fcos15-μx mg-Fsin15m =304cos15-μkmg-Fsin15m =1.3 m/s

Therefore, the magnitude of the initial acceleration of the crate is 1.3 m/s2.

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