# Question Video: Finding the Normal Reaction and the Resistance Force on a Body Moving with Uniform Velocity on a Horizontal Plane under the Action of an Inclined Force Mathematics

A body of mass 20 kg is pulled along a horizontal plane by a rope that makes an angle ๐ with the plane, where tan ๐ = 5/12. When the tension in the rope is 91 N, the body moves with uniform velocity. Find the total resistance to motion, ๐น, and the normal reaction, ๐. Use ๐ = 9.8 m/sยฒ.

05:33

### Video Transcript

A body of mass 20 kilograms is pulled along a horizontal plane by a rope that makes an angle ๐ with the plane, where tan of ๐ is equal to five twelfths. When the tension in the rope is 91 newtons, the body moves with uniform acceleration. Find the total resistance to motion ๐น and the normal reaction ๐. Use ๐ is equal to 9.8 meters per square second.

Letโs begin by sketching this out. The body has a mass of 20 kilograms, and so this means it exerts a downward force of 20๐ on the plane. Itโs pulled by a rope that makes an angle ๐ with the plane. And then, weโre told that when the tension is 91 newtons, the body moves with uniform velocity. So, the force thatโs actually pulling the body is 91 newtons.

Now, actually, there is another force that weโre interested in, and itโs a little bit outside the scope of this video to investigate this too much. But Newtonโs third law of motion tells us that for every action, thereโs an equal and opposite reaction. So, thereโs a normal reaction force of the plane on the body. Thatโs a result of the force of the weight of the body on the plane. And that acts upwards and away from the plane, as shown. Finally, letโs add the resistance to motion ๐น. We can assume that this acts parallel to the plane, as shown. This might be, say, a frictional or air resistance force.

Now, we have all the forces in our diagram. And weโre told that the body is moving with uniform velocity. Now, Newtonโs first law of motion tells us that for this to be the case, the net sum of the forces in both the horizontal and vertical direction must be equal to zero. So, weโre going to need to compare forces in the horizontal and vertical direction. This does mean, though, that we need to be really careful with the tension force thatโs acting at an angle. And so, if we add a right-angled triangle as shown, we see that there are components of this force that act in both the horizontal and the vertical direction.

We, therefore, need to use right-angled trigonometry to find those components. The hypotenuse of this triangle is 91 newtons. And then, the component that acts in a vertical direction is the opposite side. And the horizontal direction is the adjacent side. And so, weโll begin by considering the forces that act in a horizontal direction. Letโs define the adjacent side in our right-angled triangle to be ๐ฅ newtons. If we then take the direction to the right to be positive, we can say that the sum of the forces acting in this direction are ๐ฅ minus ๐น.

Then, since the body moves with uniform velocity, we can say that the sum of these forces is equal to zero. When we solve for ๐น by adding ๐น to both sides, we find ๐น is equal to ๐ฅ. Weโre actually able to work out the value of ๐ฅ by using the cosine ratio, since we know the hypotenuse and weโre trying to find the adjacent. We can say that cos of ๐ is ๐ฅ divided by 91. So, multiplying by 91 gives us ๐ฅ equals 91 cos ๐.

But we havenโt yet used the fact that tan of ๐ is five twelfths. And so, since tan of ๐ is opposite over adjacent, we can construct a more general triangle. In this triangle, the length of the side opposite to ๐ is five units and its side adjacent is 12 units. We have the Pythagorean triple five squared plus 12 squared equals 13 squared, and so, the hypotenuse must be 13. And so, cos of ๐ for our angle which is adjacent over hypotenuse must be 12 over 13. And so, ๐ฅ is 91 times 12 over 13, and thatโs equal to 84. Since ๐น is equal to ๐ฅ, we can say that ๐น must also be equal to 84. And all our measurements are in newtons, so ๐น is 84 newtons.

Weโll need to perform a similar process, but this time in a vertical direction. And that will allow us to calculate the value of ๐. Weโre going to define upwards to be positive. And weโre also going to say that the length of the side in our right-angled triangle thatโs opposite the angle ๐ is equal to ๐ฆ. This is also acting upwards. So, in an upwards direction, we have ๐ plus ๐ฆ. And then, we have 20๐ acting in the opposite direction. So, the sum of our forces is ๐ plus ๐ฆ minus 20๐. And once again, that must be equal to zero. Weโre going to add 20๐ to both sides of this equation and subtract ๐ฆ, and we get ๐ is equal to 20๐ minus ๐ฆ.

But we now need to work out the value of ๐ฆ. And so, once again, weโre going to use right-angled trigonometry. This time, we use the sin ratio since sine is opposite over hypotenuse. So, sin ๐ is ๐ฆ over 91. And so, solving for ๐ฆ, we get ๐ฆ is 91 sin ๐. But letโs go back to our more general triangle. We know the opposite side in this triangle is five and its hypotenuse is 13. And so, sin of ๐ must be five thirteenths and ๐ฆ is 91 times five thirteenths. And thatโs equal to 35.

Our earlier equation, therefore, becomes ๐ equals 20๐ minus 35. But of course, we were told that ๐ is 9.8. So, this becomes 20 times 9.8 minus 35, and thatโs 161 or 161 newtons. The resistance to motion ๐น is then 84 newtons, and the normal reaction ๐ is 161 newtons.