Question Video: Finding the Magnitude and Direction of Friction of a Body Resting in Equilibrium on an Inclined Rough Surface While an Inclined Force Is Acting on It | Nagwa Question Video: Finding the Magnitude and Direction of Friction of a Body Resting in Equilibrium on an Inclined Rough Surface While an Inclined Force Is Acting on It | Nagwa

# Question Video: Finding the Magnitude and Direction of Friction of a Body Resting in Equilibrium on an Inclined Rough Surface While an Inclined Force Is Acting on It Mathematics • Third Year of Secondary School

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A body weighing 92 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)/5. A force of 60 N is acting on the body up the line of greatest slope of the plane causing it to be in a state of equilibrium. Determine the magnitude of the friction stating whether it is acting up or down the plane, and state whether the body is on the point of moving or not.

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### Video Transcript

A body weighing 92 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 five. A force of 60 newtons is acting on the body up the line of greatest slope of the plane causing it to be in a state of equilibrium. Determine the magnitude of the friction stating whether it is acting up or down the plane, and state whether the body is on the point of moving or not.

There is an awful lot of information here. So, we’ll simply begin by sketching a diagram. We have a plane which is inclined at an angle of 30 degrees to the horizontal. The body that rests on this plane weighs 92 newtons. In other words, the force that the body exerts in a downwards direction on the plane is 92 newtons. We’re told that there’s a force of 60 newtons acting on the body up the line of greatest slope of the plane. So, that’s this force shown, and we’re also told that the body is in a state of equilibrium. In other words, the vector sum total of its forces is equal to zero.

Now, we haven’t yet taken then into account that the body rests on a rough plane. And we’re told the coefficient of friction between the body and this plane is root three over five. Now, the question is asking us to determine the magnitude of the friction, but we don’t know which direction friction is acting on the body. Let’s assume then it’s acting in the opposite direction of the force pushing the body up the plane, remembering, of course, that friction is 𝜇𝑅. 𝜇 is the coefficient of friction, and 𝑅 is the reaction force. This reaction force, 𝑅, acts perpendicular to and away from the plane as show.

Now, the purpose of drawing this diagram is to allow us to resolve forces parallel and perpendicular to the plane. We generally begin by resolving forces perpendicular to the plane. In doing so, we find the value or, at least, an expression for 𝑅. In this case though, we’re really not interested in calculating the value of 𝑅. So, we’re going to move straight on to resolving forces parallel to the plane.

We’ve already seen that we have a force of 60 newtons and a frictional force which act parallel to the plane. But we also need to take into account that the weight of the body doesn’t act parallel or perpendicular to the plane. And so, we add this right-angled triangle and resolve it into the forces which are parallel and perpendicular to the plane. Let’s call the side that we’re interested in, the part of the force which acts parallel to the plane, 𝑥. Then, we can label our triangle using the conventions for right angle trigonometry.

We’re looking to find the side opposite the included angle, and we know that the hypotenuse is 92 newtons. Since the sine ratio is sin 𝜃 is opposite over hypotenuse. We can say that sin 30 is 𝑥 over 92. And if we multiply both sides by 92, we find 𝑥 is equal to 92 sin 30. And in fact, sin of 30 degrees is one-half. So, we get 92 times a half, which is 46 or 46 newtons. Now, we were told that the body is in a state of equilibrium. This means that the sum total of the forces acting up and down the plane and parallel to the plane must be equal to zero.

An alternative way to say this is that the forces acting up and parallel to the plane must be equal to the forces acting down and parallel to the plane. So, we can either say that 60 minus the frictional force minus 46 is equal to zero or that 60 equals the frictional force plus 46. Now, of course, these are equivalent statements. We move from the first to the second by adding the frictional force and 46 to both sides of our equation.

To find the value of our friction, we simply subtract 46 from both sides. And we get friction is 60 minus 46, which is 14 or 14 newtons. Now, since this value is positive, it is indeed acting down and parallel to the plane. Remember, we assumed that in the first case, but had this been negative, that will be an indication that it was actually acting in the opposite direction. So, we have calculated the magnitude of the friction, and we’ve decided that that’s acting down the plane.

The final part of this question asks us to decide whether the body is in the point of moving or not. If it is on the point of moving, we say it is in limiting equilibrium. This is when the frictional force is at its maximum possible value before the body starts to move. In order to establish whether this is the case, we’re going to go back and look at forces perpendicular to the plane. This will now allow us to calculate the value of 𝑅 and therefore the maximum possible frictional force that the body can accept.

In order to resolve forces perpendicular to the plane, we go back to our right angle triangle. We’re now interested in calculating the value of 𝑦. This time, that’s the adjacent side in our triangle. So, we can use the cosine ratio such that cos 30 is 𝑦 over 92. And if we multiply both sides by 92, we find 𝑦 is 92 cos 30, which is 46 root three newtons. Since the body is in equilibrium, we can say that the force acting perpendicular and up from the plane must be equal to the force acting perpendicular and down from the plane. So, 𝑅 must be equal to 46 root three.

And this means the maximum possible frictional force will be root three over five, that’s the coefficient of friction, times that reaction force, times 46 root three. That gives us 27.6 newtons. The maximum possible frictional force that the body can take before it starts to move is 27.6 newtons. Remember, we calculated that, at the moment, the frictional force is just 14 newtons. And so, we say that the body is not on the point of moving.

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