8.6  Force

One of the major applications of vectors is in physics. Here are some standard examples of physical quantities that are vectors.

In this section, we consider some problems involving forces and apply vectors to solve them. In the next section we consider some problems involving velocity.

If several forces act on an object then the vector sum of them is called the resultant force. The resultant force lets you replace all the forces by one equivalent force. Here is an example.

Example: An object is acted upon by three forces, F1 = 2i − 3j, F2 = −4i + 5j, and F3 = i + j. Determine the resultant force.
Solution: Let F denote the resultant force. Then F = F1 + F2 + F3 = −i + 3j. The diagram on the left shows the three given forces. The diagram on the right shows the resultant force. This is the single force that is equivalent to the other three forces.
An object is in equilibrium if the resultant force is zero. In the previous example, for the object to be in equilibrium, we would need to apply a force to balance out the resultant. That is, a force which is the opposite of the resultant. The force we would need to apply to obtain equilibrium is i − 3j.

In this next exercise, you are asked to find the reslutant force acting on an object. Select "Level 0" for problems involving two vectors or "Level 1" for problems involving three vectors. If you have trouble recognizing the arrows as vectors, then press "Help".

Resultant Force

The resultant is the equivalent single force acting on an object. Conversely, we may decompose (or resolve) a force into the sum of two or more forces whose vector sum is the original force. The rectangular form of a vector gives a natural way to decompose a force. For example, if F = 2i + 3j, then it is the sum of F1 = 2i and F2 = 3j. We think of these as the horizontal and vertical components of the force, respectively. However, we may decompose a force into any two (or more) component forces.

Example: A 3,000 pound car rests on a hill inclined to the horizontal at 10o. What force must be applied to the car to stop it sliding down the hill.
Solution: The exaggerated diagram shows the situation. The 3,000 pound force is the gravitational force exerted by the earth. We have decomposed this force into a force F which is parallel to the incline of the hill and a force N which is perpendicular to the incline of the hill. The force N is the force exerted by the car on the road. It does not contribute to the car sliding down the hill. The force F is the force that tends to make the car slide down the hill. To prevent sliding, the opposite of F must be applied to the car. Our task is then to find F. The direction of F is clear. Note that the angle between F and the gravitational force is 80o. The magnitude of F is then 3000cos(80o) lbs or 521 lbs, approximately. So a force of 521 lbs must be applied in a direction opposite that of F to prevent sliding.

Try the following exercise. As in the preceding example, you will be asked to find either the force required to prevent the car from sliding down the hill or the force exerted by the car on the road. Give your answer numerically rounded to two decimal places.

Car on a Hill

The next example is slightly more complicated. This time we resolve the vectors vertically and horizontally. The example is completely explored in the Problems for this section and here we give only a barebone solution.

Example: An object is suspended on a wire as shown. The angles made by the wire and the horizontal support a and b as indicated. The gravitational force on the object is 1,000 newtons. Find the forces F1 and F2 acting on the wire.
Solution: It is convenient to introduce a coordinate system with the origin at the location of the object. We can now represent the gravitational force as the vector −1000j. We also write F1 and F2 in trigonometric form. Verify that
F1 = |F1|(−cos(a)i + sin(a)j)(1)
F2 = |F2|(cos(b)i + sin(b)j)(2)
The forces acting on the object are in equilibrium (the object is stationary). So F1 + F2 − 1000j = 0. This can only happen if the i components sum to zero and the j components sum to zero. Therefore, looking at the i components we get
−|F1|cos(a) + |F2|cos(b) = 0.
This gives
|F1| = |F2|cos(b)/cos(a).

From the j component we get |F1|sin(a) + |F2|sin(b) = 1000. Substitute for |F1| and solve the resulting equation for |F2| to get
|F2| = 1000cos(a)/sin(a + b).
We can now substitute back to find
|F1| = 1000cos(b)/sin(a+b).
Finally, using equations (1) and (2) we can F1 and F2.

Try this type of problem for yourself. In the next exercise set, you will be asked for the magnitude of one of the forces in the wire supporting the object. Use the equations we developed in the preceding example, but remember to replace the 1000 N with the force provided in the question. Give your answer rounded to two decimal places and provide the units of force. For example, if you work out the requested force to be 120.35738 lbs then enter 120.36 lbs as the answer.

Weight on a Wire

Note to Reviewers

We include a sample set of questions to be included in the Student Guide. These are based on the exercises in the applets and the content of each section. They are intended to be used for homework assignments and for the student to work on paper.

Click here to download the sample problems for this section.