How to Calculate Force Vectors

The main calculation of force vectors in introductory physics courses involves the decomposition of a force vector into perpendicular components. Vector addition entails placing vectors in a series, matching the head of one vector to the tail of the next, to form a chain. Such a chain is equivalent to a single vector drawn from the vector tail at one end of the chain to the vector head at the opposite end of the series. Therefore, vector decomposition into perpendicular vectors is grounded in--and equivalent to--vector addition.

Instructions

1. Suppose you have a block resting on an incline. The frictional force between the block and incline is parallel to the inclined surface. It is a function of the normal force of the block on the incline, caused by gravity. For the sake of an example, suppose that the incline is 30 degrees from horizontal and the block weighs 0.5kg.

2. Visualize (or draw) the gravitational force on the block as a vertical vector pointing down from the center of the block.

3. Draw two smaller vectors to form a right triangle with the vertical vector. Make one vector parallel to the incline, with its tail touching the tail of the vertical vector. Make the third vector perpendicular to the second vector, so that its tail touches the head of second vector, and the head touches the head of the vertical vector. Make the lengths of these two vectors such that they form a right angle where they intersect with each other. By vector addition, they sum up to equal the larger, vertical vector.

4. Have the length of the vertical vector correspond to the magnitude of the force it represent. Since 0.5kg times the gravitational constant ("g") is the magnitude of the vertical force, the vector’s length corresponds to 0.5 x 9.80 = 4.90 Newtons (N).

5. Note that the incline’s angle is the same as the angle where the normal vector intersects the vertical vector. Mark this angle between the two vectors as 30 degrees. With the vertical vector’s magnitude equaling 4.9N, the length of the tangential vector must be 4.90 x sin 30 N = 2.45 N. The length of the normal vector must be 4.90 x cos 30 N = 4.24 N.

6. Note that, as stated by the Pythagorean theorem, the squares of the two smaller vectors equals the square of the vertical vector: 4.24^2 + 2.45^2 = 4.90^2, where the caret "^" indicates exponentiation.

7. Calculate the tangential frictional force by multiplying the normal force (4.24 N) by the coefficient of friction, which for now you can suppose is 0.05. Therefore, the force of friction acts to keep the block in place with a force of 0.21 N.

8. Determine the acceleration of the block down the incline by vector addition. The 0.21 N in tangential frictional force opposes the 2.45 N in tangential gravitational force, so the effective force pulling the block down the incline is 2.24 N. Use Newton’s second law (F = ma) to divide out the block’s mass to get its acceleration along the incline’s surface: 2.24N / 0.5kg = 4.48m per second-squared.

References

• Georgia State University: Basic Vector Operations
• Cal Poly Pomona: Newton’s Second Law
• Johns Hopkins University: Cart of Frictionless Incline