Question Video: Finding the Normal Reaction of an Inclined Plane on a Body with a Force Acting on It Mathematics

A body of mass 9 kg was moving along the line of greatest slope of a smooth plane inclined at 60ยฐ to the horizontal. A force of 3 kg-wt was acting on the body, where the line of action of the force was directed toward the plane at 30ยฐ upward of the horizontal. Find the magnitude of the normal reaction of the plane on the body. Take ๐‘” = 9.8 m/sยฒ.

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Video Transcript

A body of mass nine kilograms was moving along the line of greatest slope of a smooth plane inclined at 60 degrees to the horizontal. A force of three kilogram weight was acting on the body, where the line of action of the force was directed toward the plane at 30 degrees upward of the horizontal. Find the magnitude of the normal reaction of the plane on the body. Take ๐‘” to be equal to 9.8 meters per square second.

To answer this question, weโ€™re going to begin by drawing a diagram. A body of mass nine kilograms is moving along a smooth plain inclined to the horizontal. This means, firstly, that there is a downwards force of the body on the plane. Normally, weโ€™d calculate that force to be mass times gravity. Thatโ€™s nine ๐‘” or nine ๐‘” newtons. But then, we notice we have a second force. This force acts upon the body and its units are kilogram weight. And so, we recall that one kilogram weight is roughly equal to 9.8 newtons.

Since ๐‘” is equal to 9.8 meters per square second, if we multiply both sides of this by nine, we find that nine kilogram weight is equal to nine ๐‘” or nine times 9.8 newtons. And so, we can alternatively write the downward force of the body on the plane as nine kilogram weight. Letโ€™s add everything else we know to our diagram. We know the plane is inclined at 60 degrees to the horizontal and that the line of action of the force is 30 degrees upward of the horizontal. We want to find the magnitude of the normal reaction of the plane on the body. That force acts perpendicular to and away from the plane, as shown.

Weโ€™re also told that the plane is smooth, so we know that there are no frictional forces that we need to consider. So, what do we do next? Now, we know that in the direction perpendicular to the plane, the body must be in equilibrium. Itโ€™s not moving off the plane, for example. So, the sum of the forces in this direction must be equal to zero. And so, what weโ€™re going to do next is resolve forces perpendicular to the plane. We have ๐‘… acting upwards and away from the plane, so letโ€™s take this direction to be positive. Then, we have two separate forces that we need to consider acting in the opposite direction. These are the components of our weight force and our other force that are acting perpendicular to the plane.

In order to find the components of these forces that are perpendicular to the plane, we add right-angled triangles as shown. If we look at this triangle, the one for the weight force, we know the included angle is 60. Letโ€™s label the side weโ€™re trying to find ๐‘ฅ. And we see we now have an included angle with an adjacent side of ๐‘ฅ and a hypotenuse of nine. The cosine ratio links the adjacent side with the hypotenuse such that cos ๐œƒ is adjacent over hypotenuse.

So, here, we can say cos 60 is ๐‘ฅ over nine, which means ๐‘ฅ is nine times cos 60. Since cos 60 is a half, we can rewrite this as nine times one-half which is 4.5. And so, the component of the weight force that acts perpendicular to the plane is 4.5 kilogram weight. Since this is acting in the opposite direction of our normal reaction force, we subtract this from ๐‘….

Letโ€™s repeat this process for our other force, the three kilogram weight one. Since corresponding angles are equal, we know that we can find this included angle by subtracting 30 from 60. That gives us 30 degrees. We want to find the side that Iโ€™ve labeled ๐‘ฆ or ๐‘ฆ kilogram weight. This is the opposite side in our triangle. And we know that the hypotenuse is three. This time then, we use the sine ratio. sin ๐œƒ is opposite over hypotenuse. So, sin 30 is ๐‘ฆ over three. And if we multiply by three, we find ๐‘ฆ is equal to three sin 30. Now, of course, sin 30 is the same as cos 60. Itโ€™s one-half. So, ๐‘ฆ is three times one-half, which is 1.5.

And so, the component of this force that acts perpendicular to the plane is 1.5 kilogram weight. It acts once again in the opposite direction to ๐‘…. So, we subtract it to find the sum of our forces. And we know the overall sum of these forces is equal to zero. So, ๐‘… minus 4.5 minus 1.5 is equal to zero. We can say ๐‘… minus six is equal to zero. And then, we add six to both sides of this equation to find ๐‘… is equal to six. Weโ€™re working in kilogram weights. So, we say that the magnitude of the normal reaction of the plane on the body is six kilogram weight.

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