Question

A load of 500 N is lying on an inclined plane, whose inclination with the horizontal is 30°. If the coefficient of friction between the load and the plane is 0.4, find the minimum and maximum horizontal force, which will keep the load in equilibrium.​

Answer 1

Answer:

Explanation:

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Answer 2

Given:

The weight of the load = 500N

The angle of inclination of the plane = 30°

The coefficient of friction between the load and the plane = 0.4

To Find:

The minimum and maximum horizontal force, which will keep the load in equilibrium

Solution:

The Minimum Value of F:

We know that for the minimum value of F, the body must be at the point of sliding downwards.

Also when the body is at the point of sliding downwards, then the force F = mg(sinθ − μcosθ)

Substituting mg = 500N

sinθ = sin 30° = 1/2

μ = coefficient of friction = 0.4

and cosθ = cos 30° = √3 / 2

We get F = $500 X (\frac{1}{2} - 0.4 X \frac{\sqrt{3} }{2} )$

= 500 X 0.16

= 80 N

The Minimum Value of F:

We know that for the maximum value of F, the body is at the point of sliding upwards.

Also, when the body is at the point of sliding upwards, the force F = mg(sinθ + μcosθ)

Substituting the values,

Fmax = $500X (\frac{1}{2} + 0.4X\frac{\sqrt{3} }{2} )$

= 500 X 0.84

= 420N

Hence the minimum and maximum horizontal force, which will keep the load in equilibrium are 80N and 420N respectively.