Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 19

Because the bar and the pan are in equilibrium when there are no weights, the bar and pan collectively contribute no moment about the balance point and can be left out of our equilibrium equation.

For part a), we take the moments about the balance point, which we’ll call O:

\sum M_O = W_1 a - W_2 x = 0

Solving for W_1, we get

W_1 = W_2 \frac{x}{a}

For part b), we first note that the maximum capacity will be when W_2 is slid all the way to the left end of the bar. So we use our solution to part a) and substitute in x = b = 12\:\rm{in}, a = 1\:\rm{in}, and W_2 = 1\:\rm{lb}. Thus

W_{1max} = 1\frac{12}{1} = 12\:\rm{lb}

In general, for a = 1 and W_2 = 1,

W_1 = 1 \frac{x}{1} = x

where, because I’ve dropped the units, x must be measured in inches and W_1 in pounds. We see, then, that the long arm can be marked with a linear scale, every inch representing a pound.

Problem 20Problem 18

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