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In Fig. 6-24, a forceacts on a block weighing 45N.The block is initially at rest on a plane inclined at angle θ=150to the horizontal. The positive direction of the x axis is up the plane. Between block and plane, the coefficient of static friction is μs=0.50and the coefficient of kinetic friction is μk=0.34. In unit-vector notation, what is the frictional force on the block from the plane when is (a) (-5.0N)i^, (b) (-8.0N)i^, and (c) (-15N)?

Short Answer

Expert verified

(a) The static frictional force is equivalent to the 16.6Nforce along i^.

(b) The static frictional force is equivalent to the 19.6Nforce along i^.

(c) The frictional force or kinetic frictional force acting on the block is 14.78Nalong.

Step by step solution

01

Given

Weight,W=45N

Coefficient of static friction,μs=0.50

Coefficient of kinetic friction,μk=0.34

Inclined angle,θ=150

02

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. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

Fnet=ma

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

03

Determining the free body diagram

Free body Diagram of sled over the inclined plane:

04

(a) Determining the frictional force on the block when P⏞U~=(5.0 N)(i)^

The maximum static frictional force, fmax=μsN,

By using Newton’s 2nd law of motion along vertical direction,

N=Wcos150

Thus,fmax=μsN=μs(Wcos15)=(0.50)(45)cos150=21.73N

Similarly, the maximum kinetic friction,

fmax=μkN=μk(Wcos15)=(0.34)(45)cos15=14.78N

If force P is applied along the inclination, then it’s effect on the frictional force by using Newton’s 2nd law along horizontal direction (x) is,

Fx=max

since, the block is not moving, ax=0

fs-P-Wsinθ=0fs=P+Wsinθ

Since, fsis along the opposite direction of it should be 16.6i^. This force is less than maximum static frictional force. So, the static frictional force is equivalent to the 16.6Nforce along i^

05

(b) Determining the frictional force on the block when

Similarly, here also fs<fmaxSo, the static frictional force is equivalent to the 16.6N force alongrole="math" localid="1654156238627" i^.

06

(c) Determining the frictional force on the block when 

Since, fs>fmax, then the block begins to slide. Therefore, the kinetic frictional comes to play. So, the frictional force or kinetic frictional force acting on the block is role="math" localid="1654156555096" 14.78Nalong the

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