A body of mass 16 kilograms was placed on a smooth plane that was inclined at 45 degrees to the horizontal. A horizontal force of 48 kilogram-weight was acting on the body towards the plane. Given that the line of action of the force, the body, and the line of greatest slope all lie in the same vertical plane, determine the acceleration of the body. Consider the acceleration due to gravity to be 𝑔 equals 9.8 meters per square second.
To answer this question, we’re simply going to begin by sketching a diagram. This diagram doesn’t need to be to scale, but it should be roughly in proportion so we can accurately model what’s going on. We have a smooth plane inclined at 45 degrees to the horizontal. Now, the fact that this plane is smooth tells us simply that it exerts no frictional force on the body placed on it. We have a body whose mass is 16 kilograms placed on this plane. This means it exerts a downward force on the plane. And that force is equal to mass times acceleration due to gravity. We’ll call acceleration due to gravity 𝑔, although we are told it’s 9.8 meters per square second.
The downward force the body exerts on the plane is 16𝑔. Now, we also have a horizontal force of 48 kilogram-weight acting on the body towards the plane. And so we can model this as the body exerting a horizontal force on the plane itself. The problem is we’re given information about this force in kilogram-weight. And currently, we’ve modeled the weight of the body to be in newtons. But really a kilogram-weight is simply another way of measuring a force. And we can convert from kilogram-weight to newtons by recognizing that one kilogram-weight is equal to 9.8 newtons. And so our horizontal force of 48 kilogram-weight is roughly equal to 48 times 9.8 newtons. That’s 470.4. So our horizontal force is 470.4 newtons.
We’re looking to calculate the acceleration of the body. And so we need to make an assumption about the direction in which we think the body is going to move. Let’s assume it’s going to move up the plane. And so the acceleration is positive in this direction. Really, it doesn’t matter though. If we assumed the acceleration to be acting in the opposite direction, the sign would tell us the direction in which it really acts. Now, there’s one further force that we can model on our diagram. That force is the normal reaction of the plane on the body. Let’s call that 𝑅.
Now, we don’t really need to consider 𝑅 in this question. But it’s always worth drawing on the diagram so we have everything modeled. And then the formula we’re going to use to help us to solve this problem is 𝐅 equals 𝑚𝑎: force is equal to mass times acceleration. We have modeled the acceleration as acting parallel to the plane. So we’re going to consider the components of our forces that act in this direction. Let’s begin by looking at our horizontal force. We’re going to add a right-angled triangle as shown.
We know the included angle in this triangle is 45 degrees. And that’s because corresponding angles are equal. We want to find the component of this force that is parallel to the plane. So let’s label that side of our triangle 𝑥 or 𝑥 newtons. 𝑥 is the adjacent side in this triangle, and the hypotenuse is 470.4 newtons. So let’s use the cosine ratio to form an equation. We know that cos 𝜃 is adjacent over hypotenuse. So in this case, cos of 45 degrees equals 𝑥 over 470.4. We can work out 𝑥 by multiplying by 470.4. So 𝑥 is 470.4 times cos of 45, but cos of 45 is root two over two. And so we find 𝑥 to be equal to 235.2 root two. That is the component of this force that acts parallel to the plane.
We’ll now repeat this process for the weight force. We add a right-angled triangle in. And the included angle once again is 45 degrees. I’ve called the component of this force that’s parallel to the plane 𝑦 or 𝑦 newtons. Once again, we have the adjacent and hypotenuse. So we’re going to use the cosine ratio. cos of 45 is 𝑦 divided by 16𝑔. And if we multiply by 16𝑔, we get 16𝑔 times cos of 45.
We’re now going to use the fact that 𝑔 is 9.8. Using this along the side the fact that cos of 45 is root two over two and we find 𝑦 is 78.4 root two. We now have forces acting parallel to the plane. We need to find the sum of these forces. This one acts in the same direction as the acceleration force, so we’re going to assume it’s positive. This one however acts in the opposite direction. It’s acting parallel to and down the plane. So its force is negative, meaning the sum of our forces is 235.2 root two minus 78.4 root two. This of course is equal to mass times acceleration. Remember, we’re using this formula. The mass of the body is 16 kilograms, and the acceleration we said was equal to 𝑎.
So we have an equation: 235.2 root two minus 78.4 root two equals 16𝑎. This left-hand side simplifies to 156.8 root two. So to solve our equation, our last step is to divide through by 16. 156.8 divided by 16 is 49 over five. So we find our acceleration 𝑎 is equal to 49 root two over five or 49 root two over five meters per square second. Note, now, that had we modeled the acceleration to be acting in the opposite direction — in other words, down the plane — we would have received a result of negative 49 root two over five. The magnitude of the acceleration though would have still been 49 root two over five. This just tells us that it’s decelerating in this direction.