Video: Studying the Motion of a Body Moving on a Horizontal Plane under the Action of an Inclined Force

Ed Burdette

A body weighing π‘Š is moving at a constant speed on a horizontal plane under the action of a force 145 N whose line of action is inclined at an angle πœƒ to the horizontal, where sin πœƒ = 3/5. If the resistance of the plane to the body’s motion is 1/6 π‘Š, find π‘Š and the normal reaction of the plane 𝑅.

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Video Transcript

A body weighing π‘Š 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œƒ to the horizontal, where the sin of πœƒ equals three-fifths. If the resistance of the plane to the body’s motion is one-sixth π‘Š, find π‘Š and the normal reaction of the plane 𝑅.

We can call the magnitude of the force acting on the body, 145 newtons, capital 𝐹. We’re told that the sin of the angle πœƒ involved is equal to three-fifths. And that the resistance of the plane to the body’s motion is π‘Š over six. We’ll call that resistance 𝐹 sub π‘Ÿ. We want to solve for π‘Š, the weight of the body, and also the normal reaction of the plane, called capital 𝑅.

To start out, let’s draw a diagram of this scenario. In this situation, we have a body of weight π‘Š being pushed by a force 𝐹 which is at an angle of πœƒ to the horizontal. In addition to 𝐹, 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, we’ve called that the resistive force 𝐹 sub π‘Ÿ. And there’s also the reaction force of the surface on the body. We’ve called that capital 𝑅. This is essentially the normal force. Based on our knowledge of 𝐹, the sin of πœƒ, and the resistive force 𝐹 sub π‘Ÿ, we want to solve for π‘Š and 𝑅.

To start out, we can recall Newton’s second law. This law tells us that the net force on an object is equal to the object’s mass times its acceleration π‘Ž. In our scenario, we have forces acting in two orthogonal directions. And if we define up as motion in the positive 𝑦-direction and to the right as motion in the positive π‘₯-direction, then we can write out Newton’s 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, 𝐹 sub π‘Ÿ and 𝐹. Based on Newton’s second law, we can write that 𝐹 sub π‘Ÿ minus 𝐹 times the cos of πœƒ is equal to the mass of our body multiplied by its acceleration π‘Ž.

In our problem statement, we’re told that the speed of the body is constant. Which means that its acceleration is equal to zero. Since π‘š times π‘Ž is zero, we can write that 𝐹 sub π‘Ÿ is equal to 𝐹 times the cos of πœƒ. And we recall that 𝐹 sub π‘Ÿ can be expressed as one-sixth π‘Š.

Now, let’s consider the cos πœƒ term for a moment. If we look to our diagram, we see that where the force 𝐹 is applied, we can draw that out in component form as a right triangle, where 𝐹 is the hypotenuse and 𝐹 sub π‘₯ and 𝐹 sub 𝑦 are the sides. We’re told that the sin of the angle πœƒ 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œƒ 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 it’s 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œƒ in terms of this information. Well, the sin of πœƒ is 𝑦 divided by the hypotenuse 𝐻. The cos of πœƒ is π‘₯, four, divided by the hypotenuse, five. So the cos of πœƒ is equal to four-fifths. And 𝐹, 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‘Š is equal to six times 145 newtons times four-fifths, or 696 newtons. That’s the weight force of our body.

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

We can rearrange this equation to solve for 𝑅. 𝑅 is equal to π‘Š minus 𝐹 times the sin of πœƒ. We know the sin of πœƒ. We know 𝐹. And we’ve solved for π‘Š in a previous part. So we’re ready to plug in and solve for 𝑅. Entering these values on our calculator, we find that 𝑅 is equal to 609 newtons. That’s the reaction force of the surface pushing up on the body.

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