Physics: Experiment of the Month

Mr. C’s Physics: Experiment of the Month Vector Sum of Forces

Weights are hung on the short string (m2), and on the free end of the long string (m1). The loop on the post is then adjusted up or down until the string segment running from post to knot is horizontal. This trick simplifies the analysis and clarifies the idea of force components.

The short string (m2) is vertical (up and down) and it is not moving, therefore, there must be an equal and opposite force pulling in the opposite direction. The force opposite m2 is the sum of F1 and F3.

F1 F1 + F3

Φ x F3

origin

Likewise, the junction of the origin is not moving horizontally which means there is an equal and opposite force equal in magnitude to F3.

If we measure the angle shown by Φ using a protractor, we have two parts of a right triangle. F1, the hypotenuse and the angle Φ. If we solve for the opposite side, we have F1 + F3 which equals F2, the unknown weight.

Recapping, since F3 is horizontal, the sum of the vertical components of forces F1 and F2 must be zero. For given masses m1 and m2 , we use a protractor for two purposes: 1) To ensure that F3 is horizontal, by sliding the loop until its string makes an angle of 90 degrees with the short (vertical) string that attaches to m2 . 2) To measure the angle between F1 and the horizontal (having already ensured that the looped string is horizontal).