A body of mass six kilograms was moving along the line of greatest slope of a smooth plane inclined at 60 degrees to the horizontal. A force of 28 kilogram-weight was acting on the body such that the line of action of the force was directed toward the plane at an angle of 30 degrees upwards of the horizontal. Find the magnitude of the normal reaction of the plane to the body. Take 𝑔 to be equal to 9.8 meters per square second.
Before we perform any calculations, we’re simply going to begin by sketching a diagram of this scenario. Now, these sorts of diagrams don’t need to be scale. They just should be roughly in proportion so we can accurately model what’s going on. We have a smooth plane inclined at 60 degrees to the horizontal.
Now, the fact that the plane is smooth means it won’t exert a frictional force on the body. The body itself has a mass of six kilograms. So it exerts a downwards force of mass times gravity — that’s six times 𝑔 newtons — on the plane. We then have this force of 28 kilogram-weight acting on the body toward the plane but at an angle of 30 degrees upward of the horizontal, as shown.
We do have a little bit of a problem here though. We are given this force in kilogram-weight. And we’ve modeled the downwards force of the weight of the body in newtons. And so we recall that one kilogram-weight is roughly equal to 9.8 newtons. And so we can convert this force into newtons by multiplying it by 9.8. 28 kilogram-weight is roughly 9.8 times 28, which is 274.4 newtons.
There’s one other force, and this is the normal reaction force of the plane on the body. It acts perpendicular to and away from the plane, as shown. We’re looking to find the value of this. Let’s call it 𝑅 or 𝑅 newtons. Now, we know that in the direction perpendicular to the plane, the body must be in equilibrium. It’s not moving off the plane, for example. And this means that the sum of the forces in this direction must be equal to zero.
If we let 𝐅 sub 𝑦 be a force acting perpendicular to the plane, we can say that the sum of all 𝐅 sub 𝑦 is equal to zero. So let’s find all the forces that act perpendicular to the plane. We clearly have the normal reaction force acting in one direction. We’ll take that to be positive. But what do we do about our other two forces? That’s the weight force and a force of 274.4 newtons.
Well, we’re going to need to resolve the components of these forces that act perpendicular to the plane. And so we add in right-angled triangles and use right-angle trigonometry to help us. Let’s look at the force that acts towards the plane, the 274.4-newton one. If we add in a right-angled triangle, we know that the included angle is 30 degrees. Now, this is 30 degrees because corresponding angles are equal. So the entire angle between that horizontal and the plane is 60. 60 minus 30 is 30.
We want to work out the component of this force that acts perpendicular to the plane. So this is the side of the triangle that I’ve labeled 𝑥 newtons. In fact, this side sits opposite the included angle of 30 degrees. And we know the hypotenuse of this triangle is 274.4 newtons. Since the sine ratio links the opposite and the hypotenuse, we can say that sin of 30 must be equal to 𝑥 over 274.4. Then, if we multiply both sides of this equation by 274.4, we find 𝑥 is 274.4 times sin 30. Of course, sin 30 is simply equal to a half. And so we can say that 𝑥 is 274.4 times a half, which is 137.2. And then the component of this force that acts perpendicular to the plane is 137.2 newtons.
We repeat this process with the downwards force of the body on the plane. The side that we’re trying to find, the side that’s perpendicular to the plane, let’s label 𝑦 newtons. And the included angle is 60 degrees. This time, we want to link the hypotenuse and the adjacent. So we use the cosine ratio. Cos of 60 is 𝑦 divided by six 𝑔. We’re told to take 𝑔 to be equal to 9.8. So we’re now going to replace 𝑔 with this value. And six times 9.8 is 58.8. We solve for 𝑦 by multiplying through by 58.8. And we get 𝑦 is 58.8 times cos 60 or 58.8 times one-half. That gives us 29.4. So the component of the weight that acts perpendicular to the plane is 29.4 newtons.
We’ve now calculated all of the forces that act perpendicular to the plane, with the exception of 𝑅. So let’s find their sum. We said we’d assume 𝑅 to be acting in the positive direction. This means the two values we calculated must be negative since they’re acting in the opposite direction. And so the sum of the forces that act perpendicular to the plane is 𝑅 minus 137.2 minus 29.4. And we know this is equal to zero. Negative 137.2 minus 29.4 is negative 166.6.
So we have the equation 𝑅 minus 166.6 equals zero. Let’s add this constant to both sides. That gives us 𝑅 is equal to 166.6 or 166.6 newtons. And we can therefore say that the magnitude of the normal reaction of the plane on the body is 166.6 newtons.