Video: Analysis of a Body Resting in Equilibrium on a Smooth Inclined Plane by Means of an Inclined Force

A body of weight π‘Š is placed on a smooth plane inclined at 60Β° to the horizontal. The body is kept in equilibrium under the action of a force of magnitude 54 N, which acts up the slope parallel to the line of greatest slope. Find the magnitude of the reaction 𝑅 of the plane on the body and its weight π‘Š.

05:17

Video Transcript

A body of weight π‘Š is placed on a smooth plane inclined at 60 degrees to the horizontal. The body is kept in equilibrium under the action of a force of magnitude 54 newtons, which acts up the slope parallel to the line of greatest slope. Find the magnitude of the reaction 𝑅 of the plane on the body and its weight π‘Š.

In order to answer any question of this type, it is worth drawing a diagram first. The body lies on a plane inclined at an angle of 60 degrees. Its weight π‘Š acts vertically downwards. The reaction force 𝑅 is always perpendicular to the plane. Finally, we’re told there is a force of magnitude 54 newtons acting parallel to the plane.

There are several methods we could use to solve this problem. However, as all three forces act at a point, we will use Lami’s theorem. Lami’s theorem states that if we have three forces acting at a point, in this case 𝐴, 𝐡, and 𝐢, where the angle between forces 𝐡 and 𝐢 is 𝛼, between 𝐴 and 𝐢 is 𝛽, and between 𝐴 and 𝐡 is 𝛾. Then 𝐴 over sin 𝛼 is equal to 𝐡 over sin 𝛽, which is equal to 𝐢 over sin 𝛾. You may also recognize this formula as the sine rule in trigonometry.

In our question, the angle between the weight and the plane is 30 degrees, as shown in the diagram. This means that the angle between the weight and the reaction force will be 120 degrees, as 30 plus 90 is 120. The reaction force is perpendicular to the 54-newton force. This means that the angle between them is 90 degrees.

As angles in a circle or at a point sum to 360 degrees, the angle between the weight and the 54-newton force is 150 degrees. Substituting our values into Lami’s theorem gives us 𝑅 over sin 150 is equal to 54 over sin 120, which is equal to π‘Š over sin 90. Let’s consider the first two terms. 𝑅 over sin 150 is equal to 54 over sin 120.

Whilst we could type this into the calculator, it is worth recalling some of our key trig angles. The sin of 30 degrees is equal to one-half, and the sin of 60 degrees is equal to root three over two. If two angles sum to 180 degrees, then the sine of those angles are equal. Therefore, the sin of 150 is one-half and the sin of 120 is root three over two.

We also recall from our sine graph that the sin of 90 is equal to one. We can therefore see that 𝑅 divided by one-half is equal to 54 divided by root three over two. 𝑅 divided by a half is equal to two π‘Ÿ. 54 divided by root three over two is equal to 108 over root three.

We can rationalize the denominator on the right-hand side by multiplying both the denominator and numerator by root three. Two 𝑅 is equal to 108 root three divided by three. 108 divided by three is equal to 36. So two 𝑅 is equal to 36 root three. Dividing both sides of this equation by two gives us 𝑅 is equal to 18 root three. The magnitude of the reaction 𝑅 of the plane on the body is 18 root three newtons.

Let’s now consider the first and third term in our equation. 𝑅 over sin 150 is equal to π‘Š over sin 90. Once again, on the left-hand side, we have 𝑅 divided by one-half. On the right-hand side, we have π‘Š divided by one. This simplifies to two 𝑅 is equal to π‘Š. As the reaction force 𝑅 is equal to 18 root three, then π‘Š must be equal to two multiplied by 18 root three. 18 multiplied by two is 36. So π‘Š is equal to 36 root three. We can therefore conclude that the weight of the body on the plane is 36 root three newtons.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.