Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 21

Here’s a problem where we have two unknowns, the hand force P and the reaction force at the wheel, which we’ll call R, and two equilibrium equations, one for the vertical forces and one for the moments about any point. (Because there are no horizontal forces, the horizontal equilibrium equation is 0 = 0, which, while certainly true, gives us no information.)

To avoid having to solve simultaneous equations, a common trick in problems like this is to take the moments about a point on the line of action of one of the unknown forces. The moment arm of that force will then be zero and it will drop out of the equilibrium equation. Solve the moment equation for the single unknown, and then substitute that answer into the other equilibrium equation to solve for the second unknown.

For this problem, let’s take the moments about the center of the wheel:

\sum M_O = 120\:\cdot\:15 - P\:(15 + 35) = 0

Solving, we get P = 36 lbs. Using this in the vertical equilibrium equation

\sum F_y = R - 120 + 36 = 0

we get R = 84 lbs. This is the upward force the ground exerts on the wheel. By Newton’s Third Law, the wheel then exerts a downward force on the ground of 84 lbs.

I didn’t bother drawing a free-body diagram in this case because the drawing in the book is so close to an FBD that it didn’t seem worth the bother.

Problem 22Problem 20

Last modified: January 22, 2009 at 8:32 PM.