### Video Transcript

A body weighing π is moving at a constant speed on a horizontal plane under the action of a force 145 newtons whose line of action is inclined at an angle π to the horizontal, where the sin of π equals three-fifths. If the resistance of the plane to the bodyβs motion is one-sixth π, find π and the normal reaction of the plane π
.

We can call the magnitude of the force acting on the body, 145 newtons, capital πΉ. Weβre told that the sin of the angle π involved is equal to three-fifths. And that the resistance of the plane to the bodyβs motion is π over six. Weβll call that resistance πΉ sub π. We want to solve for π, the weight of the body, and also the normal reaction of the plane, called capital π
.

To start out, letβs draw a diagram of this scenario. In this situation, we have a body of weight π being pushed by a force πΉ which is at an angle of π to the horizontal. In addition to πΉ, three other forces act on the body. There is the gravitational or weight force acting straight down. The force that resists the motion of the body to slide across the surface, weβve called that the resistive force πΉ sub π. And thereβs also the reaction force of the surface on the body. Weβve called that capital π
. This is essentially the normal force. Based on our knowledge of πΉ, the sin of π, and the resistive force πΉ sub π, we want to solve for π and π
.

To start out, we can recall Newtonβs second law. This law tells us that the net force on an object is equal to the objectβs mass times its acceleration π. In our scenario, we have forces acting in two orthogonal directions. And if we define up as motion in the positive π¦-direction and to the right as motion in the positive π₯-direction, then we can write out Newtonβs second law for both of these two directions. If we start with the π₯-direction, we can see there are two forces with components along that axis, πΉ sub π and πΉ. Based on Newtonβs second law, we can write that πΉ sub π minus πΉ times the cos of π is equal to the mass of our body multiplied by its acceleration π.

In our problem statement, weβre told that the speed of the body is constant. Which means that its acceleration is equal to zero. Since π times π is zero, we can write that πΉ sub π is equal to πΉ times the cos of π. And we recall that πΉ sub π can be expressed as one-sixth π.

Now, letβs consider the cos π term for a moment. If we look to our diagram, we see that where the force πΉ is applied, we can draw that out in component form as a right triangle, where πΉ is the hypotenuse and πΉ sub π₯ and πΉ sub π¦ are the sides. Weβre told that the sin of the angle π is equal to three divided by five. If we take away the labels of this right triangle and consider it purely from a trigonometric perspective, if the sin of π is three over five, that means that our vertical side could have a length of three and our hypotenuse have a length of five.

This triangle may start to look familiar. We can see that itβs a 3:4:5 triangle. Since we know the ratio of the horizontal side compared to the other two sides now, we can also solve for the cos of π in terms of this information. Well, the sin of π is π¦ divided by the hypotenuse π». The cos of π is π₯, four, divided by the hypotenuse, five. So the cos of π is equal to four-fifths. And πΉ, the force, is given to us as 145 newtons. Plugging in for that value and multiplying both sides of our equation by six, we find that π is equal to six times 145 newtons times four-fifths, or 696 newtons. Thatβs the weight force of our body.

Knowing that, now we want to solve for π
, which is the reaction force of the table against the body. To solve for it, weβll consider forces in the vertical or π¦-direction. And looking at our diagram, we see there are three forces with components in the vertical direction: π
, π, and πΉ. We can write that π
plus πΉ times the sin of π minus π is equal to the bodyβs mass times its acceleration, which again, because its motion is constant, we know is zero.

We can rearrange this equation to solve for π
. π
is equal to π minus πΉ times the sin of π. We know the sin of π. We know πΉ. And weβve solved for π in a previous part. So weβre ready to plug in and solve for π
. Entering these values on our calculator, we find that π
is equal to 609 newtons. Thatβs the reaction force of the surface pushing up on the body.