Question Video: Finding the Magnitude of a Force Acting on a Body Resting on an Inclined Rough Plane Given That the Body Is on the Point of Moving up | Nagwa Question Video: Finding the Magnitude of a Force Acting on a Body Resting on an Inclined Rough Plane Given That the Body Is on the Point of Moving up | Nagwa

# Question Video: Finding the Magnitude of a Force Acting on a Body Resting on an Inclined Rough Plane Given That the Body Is on the Point of Moving up Mathematics • Third Year of Secondary School

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A body of weight 438 N rests on a rough plane inclined at an angle of 30° to the horizontal. The coefficient of friction between the body and the plane is √3/2. A force acts on the body such that its line of action is parallel to the line of greatest slope. As a result, the body is on the point of moving up the plane. Determine the magnitude of the force.

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### Video Transcript

A body of weight 438 newtons rests on a rough plane inclined at an angle of 30 degrees to the horizontal. The coefficient of friction between the body and the plane is root three over two. A force acts on the body such that its line of action is parallel to the line of greatest slope. As a result, the body is on the point of moving up the plane. Determine the magnitude of the force.

We’ll begin by drawing a free-body diagram showing the body resting on a plane. We’re told that the plane is inclined at an angle of 30 degrees to the horizontal. The body itself has a weight of 438 newtons. In other words, it exerts a downwards force of 438 newtons on the plane. And of course, we know that that means that the plane exerts a normal reaction force, let’s call that 𝑅, on the body. We have a force acting on the body, let’s call that 𝐹, such that the line of action is parallel to the line of greatest slope. And then we’re told that the body is on the point of moving up the plane. Remember, this is a rough plane and that means that it exerts a frictional force on the body itself. That frictional force will be acting in the opposite direction to that which the body is trying to move. So it’s acting down and parallel to the plane as shown.

Now, we do have a little more information. We’re told the coefficient of friction. And we also know that friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the normal reaction force. So in this case, friction is going to be equal to root three over two 𝑅. And that’s great because if we can find the value of 𝑅, we can work out the exact magnitude of the frictional force. So what we’ll do is resolve forces perpendicular to the plane. And this should give us a value or an expression for the normal reaction force. The body is in limiting equilibrium, which means that the vector sum of its forces is equal to zero. We can resolve this into the various components such that the sum of the forces acting perpendicular to the plane will be equal to zero.

Now, let’s take upwards and away from the plane to be positive, and we have the normal reaction force acting in this direction. Acting in the opposite direction, we’ll need to calculate the component of the weight. So we draw a right triangle which has an included angle of 30 degrees. The component of this force that we’re trying to find is essentially the size or the magnitude of the adjacent side. And we know the hypotenuse is 438 newtons, so the adjacent side is 438 cos of 30. And so the sum of the forces acting perpendicularly to the plane is 𝑅 minus 438 cos 30. This, of course, is equal to zero. So 𝑅 minus 438 cos 30 equals zero or 𝑅 equals 438 cos 30. But of course, cos of 30 degrees is root three over two. So this is 438 root three over two or 219 root three. And so we have the normal reaction force. It’s 219 root three newtons.

And remember, this is really useful because we can now find the value of the frictional force. It’s root three over two times 219 root three. That’s 328.5 newtons. And this is great cause we’re now ready to resolve forces parallel to the plane. Once again, the sum of these forces is going to be equal to zero. And we’ll take moving up the plane as being the positive direction. So we have the force, which we’ve said is 𝐹 newtons acting upwards, and then the frictional force acting downwards. So the sum of these two forces is 𝐹 minus 𝐹 sub 𝑅. But that’s not all. We need to consider the component of the weight force that acts parallel to the plane. That’s the opposite side in our triangle. So it must be 438 sin of 30.

Now, this weight force is trying to pull the object down the plane, so we’re going to subtract this. And so the sum of the forces acting parallel to the plane is 𝐹 minus 𝐹 sub 𝑅 minus 438 sin 30. And of course, this in fact is going to be equal to zero. And we know the value of the frictional force. It’s 328.5. So we get zero equals 𝐹 minus 328.5 minus 438 sin 30. Now, sin of 30 is one-half, so 438 sin 30 is 219. Let’s solve for 𝐹 by adding 328.5 and 219 to both sides. That gives us 𝐹 equals 547.5. And so the magnitude of our force is 547.5 newtons.

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