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# Lesson 8: Free Body Diagrams

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As I make my slow pilgrimage through the world, a certain sense of beautiful mystery seems to gather and grow.

A. C. Benson, From a College Window

You already have some understanding of forces and the effect that they have on the motion of an object. For instance, you know that if you apply a large enough force to a textbook, it will move across the desk. From experience, you understand even more complicated situations, like if you apply a force to a wagon you can pull it up a hill, but if you stop pulling and let go, it will roll back to the bottom of the hill.

In order to accurately predict the motion of an object, we need to be able to determine the overall effect that several forces all acting on the same object will have on that object’s motion. We are able to do this will the use of free body diagrams (FBD).
A free body diagram is a diagram of an object, separated from its environment, with only the forces acting on it. The forces are shown as arrows that point in the direction in which the force is acting. The length of the arrow is proportional to the magnitude of the force. An example, showing all the forces acting on a moving car can be seen below:

#### Finding the net force

The net force is an imaginary force that, if applied to the object, would have the same effect as all of the other forces combined. We can determine the net force using vector addition. When trying the problem below, remember to draw your diagram to scale and add the vectors tip-to-tail.

#### Example 1

A box is pushed across the floor with a force of 100 N. If there is a 40 N force of friction between the floor and the bottom of the box, what is the net force acting on the box?

As with all problems, you must begin by stating the variables that you have been given and what variable you are required to find. This can all be done in the diagram.

Given:

Fapplied = 100 N [R];

Ffriction = 40 N [L];

Since the forces are acting in opposing directions, the problem can be solved algebraically. Let all forces acting to the right be positive and all forces acting to the left be negative.

Fapplied = 100 N [R] = 100 N

Ffriction = 40 N [L] = – 40 N

Required:

Fnet = ?

The net force (Fnet) is the force we get when we add up all of the force vectors acting on the object. We can solve to find the net force by using the vector diagram below.

Solution:

The net force acting on the box is 60 N [R].

#### Example 2

A paddler in a canoe paddles with a force of 300 N [S] while the current of the water exerts a force of 100 N [S340 °W] on the canoe. Determine the net force acting on the canoe.

As with all problems, begin by stating what is given and what you are required to find. This can all be done in the diagram.

Required: Fnet = ?
Using a scale diagram and vector addition to add the vectors tip-to-tail, we find that the overall net force acting on the canoe is 389 N [S7.4 °W].

#### Practice Questions

1. What is the net force acting on the box in the diagram below?

Given: all in the diagram

Required: Fnet

Solution:

Fnet = F1 + F2 + F3

This problem can be solved algebraically because it has forces acting in opposing directions (left and right). Assign one direction to be positive (right) and the other direction to be negative (left). When adding the vectors, all vectors going to the right will be positive and all of the vectors going to the left will be negative.

Fnet = F1 + F2 + F3

Fnet = 120 N + 60 N – 140 N

Fnet = 40 N [R]

The final answer is positive. Therefore it must be going to the right.

The net force acting on the box is 40 N [R].

2. The net force acting on the box below is 12 N [R]. Use the information provided in the diagram to determine the magnitude and direction of F1.

Given: Fnet = 12 N [R], and the diagram

Required: F1

Solution:

In this case, all forces to the right are positive.

Fnet = F1 + F2 + F3

12 = F1 + 9 N – 25 N

Always assume that the unknown force is acting in the positive direction. If this is not correct, our answer will be negative and we will know to change the direction.

12 N – 9 N + 25 N = F1

F= 28 N [R]

The magnitude and direction of Fis 28 N [R].