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A toy chest and its contents have a combined weight of 180NA toy chest and its contents have a combined weight of 0.42.The child in Fig. 6-35 attempts to move the chest across the floor by pulling on an attached rope. (a) If θis 42°whatis the magnitude of the force Fthat the child must exert on the rope to put the chest on the verge ofmoving? (b) Write an expression for the magnituderequired to put the chest on the verge of moving as a function of the angle θ. Determine (c) the value of θfor which Fis a minimum and (d) that minimum magnitude.

Short Answer

Expert verified

a)The magnitude of the force exerted by child is 74N

b)The expression for the magnitude of F or T applied by the child have been found.

c) The θ at which F or T is minimum is 23°

d) The minimum magnitude of Fis 70N

Step by step solution

01

Given

Weight, W=180N

Coefficient of static friction between the toy chest and floor, μs=0.42

Angle,θ=42°

02

Determining the concept

To find the mass of the block of C and the acceleration of block A, draw the free body diagram for each block and then apply Newton's 2nd law of equation.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 the toy chest:

04

(a) Determining the magnitude of the force exerted by child

Toy chest is moving with constant velocity that means it has 0 acceleration.By using Newton’s 2nd law of motion along the vertical direction (along y) to the system,(positive x axis along the incline and the positive y along the vertical direction),

N+Fsin42-W=0N=W-Fsin42

Thus, the frictional force is,

fs=μsN=μsW-Fsin42

Now, applying Newton’s 2nd along horizontal direction to the Block(A+C) having mass m,

localid="1654577455841" Fcos42-fs=0Fcos42-μsW-Tsin42=0Fcos42+μssin42-μsW=0F=μsW(cos42+μssin42=0.42180cos42+0.42sin42=74N

Hence, the magnitude of the force exerted by child is 74N

05

(b) Determining the expression for the magnitude of F or T applied by the child

In this case, consider the angle made by the rope to toy chest as θ, then, from above calculations,

F=μsWcosθ+μssinθ=0.42180cosθ+0.42sinθ

Hence, the expression for the magnitude of F or T applied by the child have been found.

06

(c) Determining the θat which For T is minimum

Minimize the above expression for F by working through the condition,

dFdθ=ddθμsWcosθ+μssinθ=μsWsinθ-μscosθcosθ+μssinθ2=0sinθ-μscosθ=0sinθ=μscosθsinθcosθ=μstanθ=μsθ=tan-10.42=23°

Hence, theθat which F or T is minimum is 23°

07

(d) Determining the minimum magnitude of F

Putθ=23°,F=μsWcosθ+μssinθ=0.42180cos23+0.42sin23=70N

Hence, the minimum magnitude of F is 70N

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