# physics

posted by on .

Two window washers, Bob and Joe, are on a 3 m long, 345 N scaffold supported by two cables attached to its ends. Bob, who weighs 750 N stands 1 m from the left end, as shown in Figure 8.37. Two meters from the left end is the 500 N washing equipment. Joe is .5 m from the right end and weighs 1000 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable?

I got the wrong answer.
(1m)(750N)+(2m)(500N)+(3m)(1000N)+(1.5m)(345N)=5267.5/3m=1756 N
(750N+500N+1000N+345N)-1756N=839N

The answer is suppose to be 1590N and 1005N.

• physics - ,

In your first equation, the distance to the 1000N person (Joe) from the left cable is 2.5 m, not 3 m. The term on the right side of the equation should be F2 * 3 m, where F2 is the tension in the right cable. You divide the sum of momentws on the left by 3m to get the value of F2. Your equation is also wrong becasue you have moments on the left side and a force on the right side. They have to have the same units

### Answer This Question

 First Name: School Subject: Answer:

### Related Questions

More Related Questions

Post a New Question