Motion on an Inclined Plane


Objects on inclined planes will naturally slide down the plane. Friction may play a role in preventing this at small angles but for the purpose of these examples we will neglect friction. It is also true that as the inclined plane gets steeper, the object will slide down, or accelerate at a faster rate. Problems involving inclined planes usually require determining the weight force of an object that acts parallel to the plane. This is the force that will accelerate the object.

An object that is placed on an inclined plane at an angle of \theta will have a constant weight force, w=mg downwards.

The normal force is equal in size but opposite in direction to the component of the weight force and acts at right angles to the surface. The normal force is given by: F_{ N }=mgcos\theta

The component of the weight force that acts parallel to the surface will cause the object to slide down the inclined plane. The component of the weight force that acts along the surface is given by: F=mgsin\theta

 


Example 1:

A mass of 3kg is on an inclined plane as shown in the diagram below. Determine a) the weight of the object, b) the normal force acting on the object, and c) the weight component acting parallel to the surface of the inclined plane:

a) w=mg

w=3\times 9.8

w=29.4N

b) F_{ N }=mgcos\theta

F_{ N }=3\times 9.8cos25^{ \circ }

F_{ N }=26.65N

c) F=mgsin\theta

F=3\times 9.8sin25^{ \circ }

F=12.42N


Example 2:

An object of 6.0kg mass sits on an inclined plane at an angle of 35º to the horizontal as shown below. Calculate the acceleration of the object:

First, we determine the weight component acting parallel to the surface of the inclined plane:

F=mgsin\theta

F=6.0\times 9.8sin35^{ \circ }

F=33.73N

Next we apply Newton’s 2nd Law equation, 

F=ma

33.73=6.0a

a=\cfrac { 33.73 }{ 6.0 }

a=5.62\cfrac { m }{ s^{ 2 } } down the plane

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