To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 9

This is a classic sort of problem, an academic exercise that isn’t really based on a real-world example. You’re likely to see problems like this in textbooks today.

We start with the point from which the 100 lb weight is hung. The free-body diagram of this point looks like this

The equilibrium equation for the y-direction is

\sum F_y = \frac{T_2}{\sqrt{2}} - 100 = 0

from which we learn that T_2 = 141.4\:\rm{lb}. Using this result in the equilibrium equation in the x-direction

\sum F_x = -T_1 + \frac{T_2}{\sqrt{2}} = 0

gives us T_1 = 100\:\rm{lb}.

We now move to the other point and draw the free-body-diagram

Equilibrium in the y-direction gives us the equation

\sum F_y = T_4 \frac{\sqrt{3}}{2} - 100 = 0

from which we get T_4 = 200/\sqrt{3} = 115.5\:\rm{lb}. Using this in the equilbrium equation for the x-direction

\sum F_x = T_3 - \frac{T_4}{2} - 100 = 0

gives us T_3 = 100/\sqrt{3} + 100 = 157.7\:\rm{lb}.

Problem 10Problem 8

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