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# Free body diagrams and forces tutorial

Key words:

Force

Acceleration

Newton's laws

Free body diagrams

QUESTION 1: "A 10 kg box is placed on the floor and pushed with a force of 100N. The friction coefficient is µ=0.5. Draw a Free Body Diagram and determine the acceleration of the box."

What is a "free body diagram" anyways? Basically, it is just a drawing that shows all the force arrows in the problem. Let's try to make one:

To find the acceleration, we need to use Newton's 2nd law:

We can split this up into two equations; one for the x-direction and one for the y-direction:

A forward force of 100N pushes the box in the positive x-direction, while friction opposes this force. The force of gravity just pulls the box straight down. A "normal force" counter acts it and a friction force prevents it from sliding.

But what the heck is the "normal force" exactly? Here's how to think about it:

Let's break out the microscope and look at the individual atoms of the box and the floor. Imagine that the atoms of the floor are connected by tiny springs. If you grab two of them and push them together, they will push back on your hands. If we place a box on top of them, they will get compressed. Therefore, they will push back on the box.

This picture also tells us why the friction force must be calculated the way it is:

A large normal force means that the box is "dug in" to the surface. Obviously this will make it harder to slide along the surface. If we push a box across a horizontal surface, then clearly it is not going to fly up into the air or fall into the ground. Thus the acceleration in the y-direction must be zero:

So the force of gravity and the normal force are equal in magnitude. Let us use this to simplify the x-expression:

So the speed of the box increases by 5m/s for every second we spend pushing it. QUESTION 2: "A box of mass, m, is hanging from the ceiling by two ropes which form a 60 degree angle with the horizontal. Determine the Tension in the ropes." To answer this question we need the following:

1. A sketch of the setup we are trying to model.

2. A free body diagram that keeps track of all the forces involved.

3. Some way of relating the various forces to each other.

4. A way to check if the result makes sense.

Let's start with 1.

1. A sketch of the setup we are trying to model.

Just by looking at the picture, we can predict that a heavier box means more tension in the ropes. By the way, "Tension" just means "The amount of force that the ropes pull on the box with". Obviously if the box is heavier, the ropes have to pull harder on it to make sure it stays floating in the air. In reality, if the box gets too heavy, the ropes will snap. Anyways, let's move on to 2.

2. A free body diagram that keeps track of all the forces involved.

We can see that there are 3 forces involved. There are the two tension forces from the two ropes that pull the box upwards and to the sides. There is also the force of gravity that pulls it straight down. Let's move on to 3. We need:

3. Some way of relating the various forces to each other.

Like before, let us apply Newton's 2nd law. We sum up all the forces:

What to do now? Well we know that the box is just hanging from the ceiling. It is not moving left or right, and it is not falling down or floating upwards. In other words, it is NOT accelerating. Therefore, we can write:

^Basically, since the box is not moving in either the x- or y-directions, the forces in the x- and y-directions must sum up to zero. From the two equations, we get:

We need some way of calculating the x- and y-components of the tensions. To do this, we notice that we can split these vectors up using cosine and sine:

Let us use this to rewrite the equations from before:

^Aha, so the two ropes experience the same tension. That of course makes sense. They are set up in exactly the same way. Let us plug this new information in to the y-equation:

NICE! The tension in the ropes is the weight of the box divided by 2 times the sine of the suspension angle. Let's consider the fourth and final step:

4. A way to check if the result makes sense.

Let's think about the original picture:

If we make the suspension angle closer to 90 degrees, the ropes will just hang straight down. In that case,

It makes sense that if we have two ropes holding up a box, then they each take on half of the total weight. If we imagine making the angle closer to 0 degrees, then we have

It also makes sense that if the ropes are pulling at a very flat angle, they need to pull with a lot of force to keep the box in place. QUESTION 3: "A box of mass m is placed on a slope with an angle, A. The friction coefficient between the box and the slope is µ. How big can we make A before the box starts to slide?"

To answer this question we need the following:

1. Some predictions about what will happen if we change µ, A and m.

2. A free body diagram that keeps track of all the forces involved.

3. Some way of determining the biggest value of A we can use.

4. A way to check if the result makes sense.

Ok, let's start with 1.

1. Some predictions about what will happen if we change µ, A and m.

First of all, let's be clear about what we want: A is the angle of the slope. If we make A small, then the box is pretty much standing on a flat, horizontal surface. In this case, it probably won't move. If we make A large (close to 90 degrees) then the box will obviously slide right off. So somewhere in between there must be some angle, A_max, where the box is almost ready to start slipping. What happens if we make µ small? Since µ tells us how much friction there is between the box and the slope, making µ small should make it very easy for the box to slip. If the slope is icy, then it doesn't need to be very steep for the box to begin moving. If we make µ large, then the box is strongly sticking to the surface. Imagine putting velcro or super-glue between the box and the slope. In this case we can make A large before the box begins to move. What happens if we change the mass, m, of the box? On one hand a bigger mass means that the box gets pulled more by gravity, so maybe a bigger m will make it slide more easily. On the other hand, a big m means a stronger friction force (try pushing a heavy desk or a couch across the room). So maybe a bigger m means sliding is harder? We can't really be sure, so let's find out! Now for task 2.

2. A free body diagram that keeps track of all the forces involved.

Looks good. We have the following forces:

What do we do now that the surface is no longer horizontal? It is actually kind of annoying that to work with the usual x- and y-coordinates. The box is not going to go straight to the right or drop down vertically. Instead, it will slide along the slope. Let us express the force of gravity in terms of the tangential and normal directions of the slope. We can do this by "breaking up" the Fg vector into a t- and an n-component:

Nice! We know that the box will not magically fall through the slope or fly directly away from it. Therefore the normal force from the surface must cancel out the force of gravity in the normal direction:

Now we should be ready to add up all the forces! Let's move on to 3.

3. Some way of determining the biggest value of A we can use.

Let us try to sum up all the forces that are available:

4. A way to check if the result makes sense.

Let's draw a graph of inverse tangent of the friction coefficient!

If we make µ really small, then the maximum angle is also very small. That makes sense! If the slope is icy, we can't tip it very much before the box begins to slide. If we make µ very large (by super-gluing the box to the slope), we can almost make it 90 degrees before it begins to slide.

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