Video: Finding the Tension in a String That Is Attached to a Body on an Inclined Rough Surface to Cause the Body to Be on the Point of Moving up

A body weighing 56 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)/6. The body is pulled upward by a string making an angle of 30° to the line of greatest slope of the plane. Determine the minimum tension in the string required to cos the body to be on the point of moving up the plane.

07:20

Video Transcript

A body weighing 56 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 six. The body is pulled upward by a string making an angle of 30 degrees to the line of greatest slope of the plane. Determine the minimum tension in the string required to cause the body to be on the point of moving up the plane.

There’s a lot of information here, so we’re going to begin by sketching a diagram. We have a plane that is inclined at an angle of 30 degrees to the horizontal. In fact, we’re told this is a rough plane, so we’re going to need to take into account friction. A body of weight 56 newtons rests on the plane. This means the downward force exerted by the body on the plane is 56 newtons. We know there will be a reaction force that acts in the opposite direction. In fact, it acts perpendicular to the plane. Let’s call that 𝑅.

We’re also told that the body is pulled upward by a piece of string. So, in fact, there’s a tension force here. And this string makes an angle of 30 degrees to the line of greatest slope of the plane. That’s as shown. Now, we’re actually looking to calculate the minimum tension in the string required to cause the body to be on the point of moving up the plane. In other words, the body is in equilibrium; the forces acting upwards and parallel to the plane must be equal to the forces acting downwards and parallel to the plane.

Now, in fact, there is one other force we’re interested in. And that’s the frictional force. This acts against the direction in which the body wants to move. So, it acts downwards and parallel to the plane. We will use the fact eventually that friction is 𝜇𝑅, where 𝜇 is the coefficient of friction, which we know to be root three over six. Once we’ve identified the relevant forces, our job is to resolve perpendicular and parallel to the plane.

We usually begin by resolving perpendicular so we can find a value for 𝑅. Before we do though, let’s add in these two right-angled triangles. By including these, we’ll be able to find the components for tension and weight that act parallel and perpendicular to the plane.

We begin by looking at the forces that act perpendicular to the plane. Now, the body rests on the plane. This means the forces acting upwards and perpendicular to the plane must be equal to the forces acting downwards and perpendicular to the plane. Another way to think about this is the vector sum of these forces is equal to zero.

So, let’s look at the forces acting upwards and perpendicular to the plane. We have 𝑅. That’s the reaction force. We’re also interested in the component of the tension which acts perpendicular to the plane. Let’s call that 𝑥. In fact, 𝑥 is the opposite side of our right-angled triangle. We’ve labeled the hypotenuse as 𝑇. And the included angle is 30 degrees. By using the sin ratio, that is, sin 𝜃 is equal to opposite over hypotenuse, we form the equation sin 30 is equal to 𝑥 over 𝑇. We multiply both sides of this equation by 𝑇. And we find 𝑥, that’s the component of the tension that acts perpendicular to the plane, is 𝑇 sin 30. It’s acting in the same direction as 𝑅. So, we add it. Now, in fact, sin 30 is one-half. So, this is a half 𝑇.

In the opposite direction and perpendicular to the plane is the component of the weight. Now, I’ve labeled that 𝑦. In this right-angled triangle, 𝑦 is the adjacent and the hypotenuse is 56 newtons. Using the cos ratio, we find cos of 30 is equal to 𝑦 over 56. So, the component for the weight that acts perpendicular to the plane must be 56 times cos 30. cos of 30 is root three over two. So, this is 28 root three. We subtract this because we know it’s acting in the opposite direction. And since the body is at rest, this is equal to zero.

Remember an equivalent way of writing this would have been to say that 𝑅 plus a half 𝑇 equals 28 root three. The sum of the forces acting up and perpendicular to the plane is equal to the sum of the forces acting down and perpendicular to the plane. Now, actually, we’re eventually looking to calculate 𝑇 for tension. So, we’re going to rearrange and make 𝑅 the subject.

𝑅 is equal to 28 root three minus a half 𝑇. And now that we have the expression for 𝑅, we’re going to resolve forces parallel to the plane. Remember, the body is on the point of moving, so the forces acting up and parallel to the plane must be equal to the forces acting down and parallel to the plane. Or alternatively, the vector sum of the forces is equal to zero. So, let’s see what’s acting up the plane.

Well, firstly, we have the component of the tension that’s parallel to the plane. I’ve relabeled the triangle and called this 𝑥. This time that’s the adjacent side in our triangle. So, cos 30 is adjacent over hypotenuse, which is 𝑥 over 𝑇. And so, the component for the tension that acts parallel and up the plane is 𝑇 cos 30 or root three 𝑇 over two. This is the only force acting upward and parallel to the plane. It’s going to be equal to the forces acting in the opposite direction and parallel to the plane.

Well, one of them is friction. But, of course, we know friction is 𝜇𝑅. We’ve also identified that 𝜇, the coefficient of friction, is root three over six. So, we have root three over two 𝑇 equals root three over six 𝑅. And actually, we can replace 𝑅 with our expression in terms of 𝑇. That’s 28 root three minus a half 𝑇.

There’s one more component that we’re interested in though. And that’s the component for the weight that acts parallel to the plane. I’ve relabeled the triangle and, this time, called that side 𝑦. Using the sin ratio now, we find sin 30 is equal to 𝑦 over 56. So, 𝑦 is equal to 56 sin 30. And in fact, sin 30 is one-half. So, we have 28. Remember, this is acting in the same direction as our frictional force.

So, we actually now have an equation in 𝑇. Let’s distribute our parentheses. We get 14 minus root three over 12𝑇 when we multiply each term in our bracket by root three over six. The right-hand side simplifies to 42 minus root three over 12. And we’re going to want to add root three over two 𝑇 and root three over 12𝑇. So, we’ll multiply the numerator and denominator of our first by six. So, we get six root three over 12𝑇. Then, when we add root three 𝑇 over 12 to both sides, we get seven root three over 12 𝑇 equals 42.

We want to solve for 𝑇, so we’ll divide both sides by seven then multiply through by 12. So, root three 𝑇 equals 72. Finally, when we divide by root three, we get 72 over root three, which simplifies to 24 root three. And we found that the tension in the string is 24 root three newtons. Now, of course, this is indeed the minimum tension in the string. If the tension was any greater, then the forces acting up and parallel to the plane would be greater than the forces acting down and parallel to the plane. And the body will begin to move upward.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.