Question Video: Analysis of the Equilibrium of an Inclined Uniform Rod Resting on a Horizontal Rough Plane and a Support Mathematics

Video Transcript

A uniform rod 𝐴𝐵 of weight 10 newtons and length 12.5 meters is resting with this end 𝐴 on a rough horizontal plane and point 𝐶, between 𝐴 and 𝐵, resting against a smooth horizontal nail, which is 5.7 meters above the horizontal plane. If the rod is about to slide when it is inclined to the horizontal at an angle whose tangent is three-quarters, determine the coefficient of friction between the rod and the horizontal plane.

We can call the coefficient of friction between the rod and the horizontal plane 𝜇. In this statement, we’re told that the rod has a weight — we can call it 𝑊 — of 10 newtons and also that its length, which we can name 𝐿, is 12 and a half meters. We’re told that this rod rests on a nail at point 𝐶, which is a distance we can call 𝑑 above the horizontal plane. 𝑑 is given as 5.7 meters.

If we call the angle between the horizontal surface and the rod 𝜃, we’re told in this statement that the tangent of 𝜃 is equal to three-fourths. And based on all this information, we want to solve for 𝜇, the coefficient of friction between the rod and the plane.

As we begin work towards our solution, notice that the rod is stationary. Despite the forces acting on it, it’s not in motion. This means we can write that the sum of all the forces acting on the rod add up to zero. And even more, the sum of all the torques that act on the rod also add up to zero. It’s these two conditions and working them out that will give us the information we need in order to solve for 𝜇.

Let’s focus first on what information we can find by using the fact that the sum of the forces on the rod is zero. We’ll start by drawing in all the forces that act on the rod while it’s in this position. First, if we put a dot at the center of the rod that is halfway along its length, then since the rod is uniform along its length, the weight force points straight down from that location.

The nail also exerts a force on the rod. And the direction of this force is perpendicular to the rod’s length. We can call this force 𝐹 sub 𝑟. At the bottom of the rod at point 𝐴, there’s a normal force we’ll call 𝐹 sub 𝑁 that points straight up. And there’s also a frictional force we can call 𝐹 sub 𝑓 that points to the right in the horizontal direction. These are all the forces that are acting on the rod.

If we define positive directions to be up and to the right, then we can write two force balance equations, one for horizontal forces and one for vertical forces acting on the rod. Before we write those equations down, we notice that three of our four forces are already pointed entirely along the horizontal or vertical directions. The only exception is 𝐹 sub 𝑟. So we’ll want to break this force into its horizontal and vertical components. If we draw a horizontal line out from point 𝐶, then the angle between that line and the rod must be equal to 𝜃.

That means that the angle between the force vector 𝐹 sub 𝑟 and that horizontal line must be equal to 90 degrees minus 𝜃. Knowing this, we can now start to write out our two force balance equations. In the vertical direction, we can write that the normal force, 𝐹 sub 𝑁, added to 𝐹 sub 𝑟 times the sin of 90 degrees minus 𝜃 minus the weight force 𝑊 is equal to zero. Because of the particular phase relationship between sin and cos of 𝜃, sin of 90 degrees minus 𝜃 is equal simply to the cos of 𝜃. Those are our forces in the vertical direction.

In the horizontal direction, we can write that 𝐹 sub 𝑓, the frictional force, minus 𝐹 sub 𝑟 times the cos of 90 degrees minus 𝜃 is equal to zero. Once again, our trigonometric function can switch based on the phase relationship between cosine and sine. So we now have two independent equations, but we have three unknowns: 𝐹 sub 𝑁, 𝐹 sub 𝑟, and 𝐹 sub 𝑓. We’ll need a third independent equation to help us solve for these values. We can find that equation by using the fact that the net torque on this rod is also equal to zero.

If we decide that rotation in the clockwise direction is positive and we also choose as our rotation point the geometric center of the rod, then we’re almost ready to start writing out our torque-equals-zero equation. Before we write that equation, we’ll want to find the perpendicular components of 𝐹 sub 𝑁 and 𝐹 sub 𝑓 that act as torque levers. And we’ll also want to solve for a distance — we can call it ℎ — between point 𝐶 and the midpoint of the rod. For the component of the normal force 𝐹 sub 𝑁, we can write that the torque due to this force is 𝐹 sub 𝑁 times the sin of 90 degrees minus 𝜃 multiplied by the length of the rod over two, 𝐿 over two.

And as before, we can rewrite this sine expression as simply cos of 𝜃. Now for the torque about the midpoint caused by 𝐹 sub 𝑟, that’s simply equal to 𝐹 sub 𝑟 times ℎ. But we want to solve for that value ℎ. We can draw a triangle that shows the information we know to help us solve for the length ℎ. Considering this right triangle, we’re told that the tangent of the angle 𝜃 is equal to three over four. And this triangle therefore must be a 3:4:5-triangle.

Looking then back at our top triangle, this must mean that the sin of the angle 𝜃, which equals three-fifths, based on our discovery of our triangle as a 3:4:5-triangle must be equal to 𝑑, the height from ground level to the point 𝐶, divided by 𝐿 over two plus ℎ. So working with this equation, we’re able to rearrange and solve for ℎ. ℎ is equal to five 𝑑 over three minus 𝐿 over two. And plugging in our given values for 𝑑 and for 𝐿, calculating this value with 𝑑 and 𝐿 plugged in, we find that ℎ is equal to thirteen-fourths meters.

This is the distance we can use to multiply 𝐹 sub 𝑟 in our torque equation. With the torques caused by 𝐹 sub 𝑁 in and 𝐹 sub 𝑟 taken care of, we now have 𝐹 sub 𝑓, the frictional force, to account for. We can write that torque as 𝐹 sub 𝑓 times the sin of 𝜃 multiplied by 𝐿 over two. And all three of these torques added together sum to zero. We now have three independent equations and three unknowns. If we solve first for 𝐹 sub 𝑟, the force that is created on the rod by the nail, then we can use our horizontal force balance equation to find a substitute for 𝐹 sub 𝑓 in terms of 𝐹 sub 𝑟 to plug in to our torque equation.

When we make this substitution, the new term becomes 𝐹 sub 𝑟 times the sin squared of 𝜃. Next we’ll look to replace 𝐹 sub 𝑁 with something in terms of 𝐹 sub 𝑟. To do that, we’ll look at our vertical force balance equation. When we rearrange this equation, we see that 𝐹 sub 𝑁 is equal to 𝑊 minus 𝐹 sub 𝑟 cos 𝜃. We can then insert this term to replace 𝐹 sub 𝑁 in our torque balance equation. When we do, we now have an expression entirely in terms of one unknown variable, 𝐹 sub 𝑟, and known constants: 𝐿, 𝜃, and 𝑊.

When we plug in 10 newtons for 𝑊 and 12.5 meters for 𝐿 and replace cos 𝜃 terms with four-fifths and sin 𝜃 terms with three-fifths, and after plugging in and solving for 𝐹 sub 𝑟, we find that it’s equal to 100 over 19 newtons. We then take this value for 𝐹 sub 𝑟 and insert it into our vertical force balance equation, plugging in also for 𝑊 which is 10 newtons and the cos of 𝜃 which is four-fifths. When we compute this difference, we find that 𝐹 sub 𝑁, the normal force, is 110 over 19 newtons.

We’re also able to use 𝐹 sub 𝑟 to solve for 𝐹 sub 𝑓, the frictional force. If we take our value for 𝐹 sub 𝑟 and substitute it into our equation, 𝐹 sub 𝑓 equals 𝐹 sub 𝑟 sin 𝜃 where sin 𝜃 can be written as the fraction three-fifths, then when we multiply these two fractions through, we find that the friction force is sixty nineteenths newtons. Now there’s one last equation for us to recall in order to find out what 𝜇 is.

In general, the force of friction 𝐹 sub 𝑓 is equal to the coefficient of friction times the normal force. And if we rearrange this expression to solve for 𝜇, we find it’s 𝐹 sub 𝑓 over 𝐹 sub 𝑁. We’ve calculated both those forces and are able to plug them in now. When we do, we see that the factors of one over 19 cancel and the expression reduces to six divided by 11. That’s the coefficient of friction that exists between the rod and the horizontal floor.