Video: Calculating Moments for a Uniform Rod

A uniform rod 𝐴𝐡 weighing 111 N is resting in a vertical plane with its upper end 𝐴 against a smooth vertical wall and its lower end 𝐡 on a rough horizontal floor. If the rod is resting in limiting equilibrium when inclined by an angle of 30Β° to the horizontal, find the coefficient of friction πœ‡ between the rod and the floor and the reaction of the wall 𝑅_(π‘Ž) at its upper end 𝐴 rounded to two decimal places.

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Video Transcript

A uniform rod 𝐴𝐡 weighing 111 newtons is resting in a vertical plane with its upper end 𝐴 against a smooth vertical wall and its lower end 𝐡 on a rough horizontal floor. If the rod is resting in limiting equilibrium when inclined by an angle of 30 degrees to the horizontal, find the coefficient of friction πœ‡ between the rod and the floor and the reaction of the wall 𝑅 sub π‘Ž at its upper end 𝐴 rounded to two decimal places.

Before we perform any calculations here, we’re going to begin by drawing a free body diagram. Remember, this is a very simple diagram that represents all the possible forces that we’re interested in. Here is our uniform rod resting with its upper end 𝐴 against the wall and the lower end 𝐡 on the horizontal floor. Now, the fact that the wall is vertical and the floor is horizontal means that they are perpendicular to one another. They meet it 90 degrees.

We’re also told that the rod is uniform and it weighs 111 newtons. This means it exerts a downwards force of 111 newtons. And because it’s uniform, we can consider that to be acting exactly halfway along the rod. We’re not actually given the length of the rod. And so we’re going to define it to be equal to two π‘₯. And let’s call that meters. This means that the downwards force of the weight of 111 newtons is acting π‘₯ meters from either end.

We’re told that the rod is inclined at an angle of 30 degrees to the horizontal. And when this happens, the rod is in limiting equilibrium. This means it’s in equilibrium, but it’s on the point of sliding. Since the floor is rough then, this means it must be exerting a frictional force on the rod itself. That frictional force will always act in the opposite direction in which the object is trying to slide. So on our diagram here, it’s acting to the right. There are, in fact, two more forces that we need to model.

We know by Newton’s third law of motion that since the rod will be exerting a force on the floor and the wall, the floor and the wall must themselves exert a normal reaction force on the rod. We model them as 𝑅 sub 𝑏 and 𝑅 sub π‘Ž, respectively. Now that we have all of the relevant forces, we can begin to perform some calculations. The calculations that we perform come from the fact that the rod is in equilibrium. And we know that for an object to be in equilibrium, the sum of all forces acting upon that object must be equal to zero.

In scenarios like this, we tend to think about these in terms of direction. We could say that the sum of the forces acting in the π‘₯-direction horizontally is equal to zero and the sum of the forces acting in the 𝑦-direction, that’s vertically, is also equal to zero. But we also know that for an object to be in equilibrium, the sum of moments of the forces must also be equal to zero, where a moment is calculated by multiplying the force by the perpendicular distance of the line of action of that force from the point about which the object is trying to turn.

So let’s begin with our first fact. That is, the sum of all of our forces is equal to zero. We’re going to resolve forces both horizontally and vertically. And in fact, we’ll begin with the vertical direction. Let’s call that 𝐹 sub 𝑦. Let’s define upwards to be the positive direction here so that the 𝑅 sub 𝑏 force is acting in the positive direction and the 111-newton force is acting in the negative direction. We can, therefore, say that the sum of the forces that act vertically on our diagram is 𝑅 sub 𝑏 minus 111. Now, of course, we said that the sum of these forces is zero. So 𝑅 sub 𝑏 minus 111 equals zero. And we can solve this equation for 𝑅 sub 𝑏 by adding 111 to both sides. And we get 𝑅 sub 𝑏 is equal to 111 or 111 newtons.

Let’s repeat this process and consider the forces acting in the horizontal direction. We’ll call that 𝐹 sub π‘₯. We’ll choose the direction in which the frictional force is acting as being positive. We can, therefore, say that the sum of the forces acting in this direction is the value of the frictional force at 𝐡 minus the reaction force at π‘Ž. And once again, that is equal to zero. Now, how does this help us? Currently, we have two unknowns. But we are able to recall a formula that will help us to calculate the value of the frictional force.

We know that the friction is calculated by multiplying πœ‡ by 𝑅. πœ‡ is the coefficient of friction. It essentially tells us how rough the object is. And 𝑅 is the normal reaction force at that point. Since the frictional force is acting at the point 𝐡 on the rod, we can say that πœ‡π‘… is πœ‡ times 𝑅 sub 𝑏. And our equation, therefore, becomes πœ‡ times 𝑅 sub 𝑏 minus 𝑅 sub π‘Ž equals zero. But remember, we calculated 𝑅 sub 𝑏 to be equal to 111 newtons. And so this equation becomes 111πœ‡ minus 𝑅 sub π‘Ž equals zero. We can solve this equation for 𝑅 sub π‘Ž by adding 𝑅 sub π‘Ž to both sides. And so we find 𝑅 sub π‘Ž is equal to 111πœ‡.

Now we’ve done nothing truly groundbreaking yet. But we can carry all of this information forward. We’re now going to think about the moments of the forces. We know that the sum of these moments is equal to zero. So let’s keep the key information on screen and clear some space.

We’re going to choose a point about which to take moments. Now, we can actually choose any point on the rod, but since there are more forces acting at 𝐡 than there are at 𝐴, we’re going to take moments about 𝐡. We’re also going to define the counterclockwise direction to be positive as shown. Moments, remember, are calculated by multiplying the value of the force by the perpendicular distance of the line of action of the force to the point about which the object is trying to rotate. So here we need to find the value of the force and its perpendicular distance from the line of action to point 𝐡.

We know that we’re not interested in any of the forces acting at 𝐡 since those are zero meters away from 𝐡. And so their moments are going to be equal to zero. So instead, we move on to the downwards force of the weight, 111 newtons. In order to find the moment of this force, we’re going to need to calculate the component of it that acts perpendicular to the rod. And so we add in a right-angled triangle as shown. The included angle in this triangle is 30 degrees. And since we want to find its component that is perpendicular to the rod, let’s call that length 𝑦.

Relative to our included angle, we then see that 𝑦 is the adjacent side. And we know that the hypotenuse is 111 newtons. So we’re going to use the cosine ratio. cos of πœƒ is adjacent over hypotenuse. Substituting what we know about our triangle into this formula, and we get cos of 30 is 𝑦 over 111. And if we multiply by 111, we find 𝑦 to be equal to 111 times cos of 30. cos of 30, of course, is root three over two. So we can further simplify this as 111 times root three over two. Now the moment is going to be negative since this force is trying to rotate the rod in a clockwise direction. Force times distance here is 111 root three over two times π‘₯.

And so now we’re going to move on to another force on our diagram. This time, we need to find the component of the reaction force at 𝐴 that is perpendicular to the rod. So we draw in another right-angled triangle. We have an included angle of 30 degrees once again. And we can convince ourselves that is true since we know alternate angles are equal. Let’s label the side of the triangle that we’re trying to find 𝑧. 𝑧 is the opposite side of this triangle. And of course, we have an expression for the hypotenuse. So this time, we use the sine ratio. sin of 30 degrees here is 𝑧 over 𝑅 sub π‘Ž. And if we multiply by 𝑅 sub π‘Ž, we get 𝑧 is 𝑅 sub π‘Ž sin 30.

Now, in fact, sin of 30 degrees is one-half. So this is a half 𝑅 sub π‘Ž. Now that we’ve calculated the component of this force that’s perpendicular to the rod, we can find its moment. It’s trying to turn the rod in a counterclockwise direction. And so the moment is going to be positive. Force times distance is a half 𝑅 sub π‘Ž times two π‘₯. And of course, we know the sum of these moments must be equal to zero. Let’s clear some space and see what else we can do.

We might notice that we currently have two unknowns in this equation. We have π‘₯ and 𝑅 sub π‘Ž. But of course, π‘₯ and two π‘₯ are dimensions; they’re lengths of our rod. And so π‘₯ cannot be equal to zero. And this means we can divide our entire equation by π‘₯. A half 𝑅 sub π‘Ž times two is just 𝑅 sub π‘Ž. And so we can solve this equation for 𝑅 sub π‘Ž by adding 111 root three over two to both sides. So 𝑅 sub π‘Ž is 111 times root three over two. That gives us 96.128 and so on. Rounding this correct to two decimal places, and we find 𝑅 sub π‘Ž is 96.13 newtons.

Now, we need to find the value of πœ‡. And so we go back to one of our earlier equations. 𝑅 sub π‘Ž is 111 times πœ‡. This means 111πœ‡ must be equal to 111 times root three over two. We solve for πœ‡ by dividing both sides by 111. So πœ‡ is root three over two, which is 0.866 and so on, or 0.87 correct to two decimal places. 𝑅 sub π‘Ž is 96.13 newtons, and πœ‡ is 0.87.

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