Video Transcript
A body of mass 10 kilograms is placed on a smooth plane inclined at 30 degrees to the horizontal. A horizontal force of magnitude 34 kilogram weight is acting on the body toward the plane. The line of action of the force, the body, and the line of greatest slope all lie in the same vertical plane. Taking 𝑔 to be equal to 9.8 meters per square second, determine the magnitude of the normal reaction of the plane on the body, rounding your answer to two decimal places.
Before we do any calculations, let’s begin by drawing the diagram. We have a body with a mass of 10 kilograms placed on a plane inclined at 30 degrees to the horizontal. We know that the downwards force the body exerts on the plane is mass times gravity. So that’s 10𝑔, and that’s in newtons. We then have this horizontal force with a magnitude 34 kilogram weight acting in the direction shown. Now, kilogram weight is a different unit to measure force, so we have newtons and kilogram weight. What we’re going to do is convert from kilogram weight into newtons.
We know that one kilogram weight is roughly 9.81. Let’s call that 9.8 newtons. And so, to convert from 34 kilogram weight into newtons, we’re going to multiply 34 by 9.8. And when we do, we find that the horizontal force is of magnitude 333.2 newtons. The question wants us to find the magnitude of the normal reaction of the plane on the body. That’s this force shown. It’s the normal reaction. So, it’s perpendicular and acting away from the plane. We know that the body is in equilibrium in a direction perpendicular to the plane. And so, we can say that the sum of all of its forces in this direction must be equal to zero.
We clearly see we have a reaction force that’s perpendicular to the plane, but we’re going to need to consider the components of this horizontal force and the weight of the body that also act in this direction. And so, we’ll begin with the component of the weight that acts perpendicular to the plane. We’re going to add a right-angled triangle. The included angle in this triangle here is 30 degrees. And since we’re trying to find the component of the weight that acts perpendicular to the plane, we want to find the value of 𝑥.
𝑥 is the adjacent side in our right-angled triangle. And we know that the hypotenuse is the main line of action of the force; it’s 10𝑔. To link the adjacent and the hypotenuse, we use the cosine ratio. So, cos 𝜃 is adjacent over hypotenuse, meaning cos 30 is 𝑥 over 10𝑔. We can find the value of 𝑥 by multiplying both sides of this equation by 10𝑔. So, 𝑥 is 10𝑔 cos 30 or 10 times 9.8 — since we’re told 𝑔 is 9.8 meters per square second — times cos 30. Since cos 30 is root three over two, we can rewrite this as 49 root three, meaning that the component of the weight that acts perpendicular to the plane is 49 root three newtons.
We’re going to repeat this with the horizontal force. We add a right-angled triangle whose included angle is once again 30 degrees. We want to find the value of 𝑦 this time. Remember, we’re looking for the component that’s perpendicular to the plane. 𝑦 is the opposite side in our triangle, and then we know that the hypotenuse is 333.2 newtons. This time, we use the sine ratio to link the opposite and the hypotenuse. sin 30 is 𝑦 over 333.2, meaning 𝑦 is 333.2 times sin 30. sin 30 is one-half. So, we see that 𝑦 is 166.6. And the component of this horizontal force that acts perpendicular to the plane then is 166.6 newtons.
Now that we’ve calculated the forces that act perpendicular to the plane, we’re going to find their sum. If we take the normal reaction force to be acting in a positive direction, then the other two forces are acting in a negative direction. And so, the sum of the forces on our body that act perpendicular to the plane is 𝑅 minus 49 root three minus 166.6. We know the body is in equilibrium, so the sum is equal to zero. We’re going to solve for 𝑅 by adding 49 root three and 166.6 to both sides.
So, 𝑅 is 49 root three plus 166.6, giving us 𝑅 is 251.4704 and so on. We’re told to round our answer to two decimal places. That’s 251.47. And so, we see the normal reaction of the plane on the body is 251.47 newtons.
