A box is dropped on a conveyor belt moving at 3.4 meters per second. If the coefficient of friction between the box in the belt is 0.27, how long will it take before the box moves without slipping?
Some important information in this problem is the speed of the conveyor belt, 3.4 meters per second, and the coefficient of friction between the box and the belt of 0.27. Also we’re asked to solve for a time which we can represent as lowercase 𝑡, which tells us how much time elapses between the time the box initially makes contact with the belt and the time that the box moves without slipping.
To get a little bit more familiar with this problem, let’s draw a diagram of what’s going on. So here we have our box being dropped onto a conveyor belt, and the belt we know is moving at 3.4 meters per second. And when the box comes in contact with the belt, there is a coefficient of friction between them of 0.27.
And we want to solve for the time value that represents the time elapsed between the box landing on the bell in the box moving with the bell without slipping. So just physically you can picture that when this box drops on the belt, at first it will slide as it tries to catch up with the speed of the belt but eventually it will grip the belt and move along with it at that same speed. And our project is to find out how long will that take.
Now let’s imagine that we advance time a little bit in this diagram so that the box is no longer in the air but it’s just made contact with the belt. Now that the box is touching the belt, it’s going to experience a frictional force due to the belt.
That frictional force we can call 𝐹 sub 𝑓, and the force on the box pushes the box in the direction of the belt, left to right. Now as we look at the forces on the box, vertically there will be the force of gravity and the normal force. But we’re interested only in horizontal forces. and in the horizontal direction, this frictional force is the only force that acts on the box.
There is nothing else pushing it forward or pulling it back to the left. So again the frictional force on the box is the only horizontal force acting on it. Now when we talk about forces, you may recall Newton’s second law.
That law states that the net force on an object is equal to the mass of that object multiplied by its acceleration. Now as we’re just considering motion in the horizontal direction, we can write a version of Newton’s second law for the horizontal forces on the box.
This expression would read that the frictional force, our one horizontal force, is equal to the mass of the box multiplied by its acceleration. Now what is frictional force?
Well, recall the definition of that force Friction force, 𝐹 sub 𝑓, is equal to the Greek letter 𝜇, which symbolises the coefficient of friction, multiplied by the normal force that acts on the object.
And the normal force we know is equal to the mass of that object times 𝑔 when the object is sitting on a horizontally flat surface. So normal force equals the mass of the box multiplied by 𝑔. We can substitute not into equation for 𝐹 sub 𝑓.
So rewriting our version of Newton’s second law for the horizontal forces on the box, we see that 𝐹 sub 𝑓 equals 𝜇 times the mass of the box times 𝑔, which is equal to the mass of the box multiplied by its acceleration.
Looking at this equation, we see that the mass cancels out. This problem is independent of the mass of the box. So here is what we found. We found that the acceleration of the box under these conditions is equal to 𝜇 times 𝑔.
Now we may be looking at that and wondering, “Okay, that’s great. But we wanted to solve for the time that it takes for the box to get up to speed with the belt and stop slipping.” Now we can use one of the keys of this problem, which is that the acceleration that the box undergoes is constant in this process.
Since we’re dealing with constant acceleration, that means that our array of kinematic equations applies to the situation. Let’s recall what those kinematic equations are. These kinematic equations again apply when we have an object moving with constant acceleration, just like we do in this problem.
Now keeping in mind that we’re talking about motion only in the horizontal direction for our problem, let’s look through these kinematic equations and see if we can find one that helps us. The one at the very top, that final velocity is equal to initial velocity plus acceleration times time, applies very well to our situation.
We know that in the horizontal direction, 𝑣 sub 𝑓 of the box is given as the speed of the belt, 3.4 meters per second, and 𝑣 sub 𝑖, the initial speed of the box in the horizontal direction is zero. That leaves us with acceleration, which we’ve just solved for as 𝜇𝑔, and time, the variable that we wanna solve for.
Let’s rewrite this equation in terms of the particulars of our problem. We get that 3.4 meters per second is equal to 𝜇 times 𝑔 times 𝑡. If we divide both sides of equation by 𝜇 times 𝑔, then both 𝜇, the coefficient of friction, and 𝑔, the acceleration due to gravity, cancel out on the right side of our equation.
We’ve isolated 𝑡, the variable we wanna solve for, and it equals 3.4 meters per second divided by 𝜇 times 𝑔. Now we’re given 𝜇 in this problem; that’s 0.27. And 𝑔 we treat as an exact number of 9.8 meters per second squared. So let’s plug those into our equation now.
With all the values plugged in, we’re ready to use our calculator to solve for the final value of 𝑡. That value is 1.3 seconds.
Notice that our answer has two significant figures, just like the values given in the problem statement. So the box takes 1.3 seconds to go from slipping against the belt to caught in traction and moving at the same speed as the belt, at 3.4 meters a second.