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Inclined Planes

An object with mass M kg rests on a plane which is inclined at an angle of θ to the the horizontal.

Since the object is at rest, the forces are in equilbrium.

The normal force, R, is perpendicular to the object.

The frictional force, F, stops the object moving down the plane.

Trig refresher - if required

Resolving Mg parallel and perpendicular to the plane gives

R = Mgcosθ

and F = Mgsinθ

The maximum value of F is

where μ is the coefficient of friction.

For equilbrium,

At the position of limiting equilibrium, the object is on the point of moving and

For the object to actually move down the plane,

If a force is applied up the plane which is greater than μR, the object will move up the plane. The frictional force, F, then acts down the plane.

For movement up the slope :

Example

An object of mass 10 kg is at rest in limiting equilibrium on a slope which is inclined at 30° to the horizontal.

1) Find the coefficient of friction between the object and the plane.

2) Find the magnitude of the forces experienced by the object at right angles to and parallel to the plane. (Take g = 9.8m/s² )

Solution

Resolving forces,

The object is in limiting equilibrium, so

Alternatively,

Example

An object of mass 1kg is placed on an inclined slope of 35° to the horizontal as shown:

Applying a force of 3N up the slope and parallel to the plane puts the object in limiting equilibrium. What is the coefficient of friction ? (Take g = 9.8m/s² )

Solution

Resolving forces:

but

Example

A block starts from rest and slides straight down a slope inclined at an angle θ to the horizontal, where sin θ = 0.25 and the coefficient of friction between the block and the slope is 0·125.

Find the speed of the block after it has travelled 5 metres.

Solution

Given sinθ=0.25