(^142) A Textbook of Engineering Mechanics

Solution. Given: Load (W ) = 300 N; Force (P 1 ) = 60 N and angle at which force is

inclined (θ) = 30°,

Let α = Angle of inclination of the plane.

First of all, consider the load lying on a smooth plane inclined at an angle (α) with the

horizontal and subjected to a force of 60 N acting at an angle 30° with the plane as shown in Fig.

8.18 (a).

Fig. 8.18.

We know that in this case, because of the smooth plane μ = 0 or φ = 0. We also know that

the force required, when the load is at the point of sliding upwards (P),

sin ( ) sin

60 300

cos ( – ) cos 30

W

α+φ α

=× = ×

θ φ °

sin

300 346.4 sin

0.866

α

=× = α

...(Q φ = 0)

or

60

sin 0.1732

346.4

α= = or α = 10°

Now consider the load lying on the rough plane inclined at an angle of 10° with the horizontal

as shown in Fig. 8.18. (b). We know that in this case, μ = 0.3 = tan φ or φ = 16.7°.

We also know that force required to move the load up the plane,

sin ( ) sin (10 16.7 )

300 N

cos cos 16.7

PW

α+φ °+ °

=× = ×

φ°

sin 26.7 0.4493

300 300 140.7 N

cos 16.7 0.9578

°

=× =× =

°

Ans.

Alternative method

Ist case

Given: In this case load (P) = 60 N; Angle (θ) = 30° and force of friction F = 0 (because of

smooth plane). Resolving the forces along the inclined plane,

60 cos 30° = 300 sin α

∴

60 cos 30 60 0.866

sin 0.1732

300 300

°×

α= = = or α = 10°

2nd case

Given: In this case, coefficient of friction (μ) = 0.3 = tan φ or φ = 16.7°

Let P = Force required to move the load up the plane,

R = Normal reaction, and

F = Force of friction equal to 0.3 R.

Resolving the forces along the plane,

P = F + 300 sin 10° = 0.3 R + (300 × 0.1732) = 0.3 R + 51.96 N ...(i)

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