Posted by **Catherine** on Monday, May 2, 2011 at 5:13pm.

A beam, B1, of mass M1 = 0.6887 kg and length L1=1 m is pivoted about its lowest point at P1. A second beam, B2, of mass M2=0.200 kg and length L2=0.200m is suspended (pivoted) from B1 at a point P2, which is a horizontal distance d=0.550, from P1. To keep the system at equilibrium, a mass m =0.500kg has to be suspended from massless string that runs horizontally from P3, at the top of beam B1, and passes over a frictionless pulley. The string runs at a vertical distance y = 0.707 m above the pivot point P1. Determine the net torque at P1, P2, and P3.

## Answer this Question

## Related Questions

- Physics - A beam of mass mb = 10.0 kg, is suspended from the ceiling by a ...
- Physics - A beam of mass mb = 10.0 kg, is suspended from the ceiling by a single...
- physics - A beam of negligible mass of length 3 m is suspended by a string ...
- Physics - A purple beam is hinged to a wall to hold up a blue sign. The beam has...
- Physics - A beam has a mass M and length L. A point mass m is located in ...
- physics - In Fig. 12-51, uniform beams A and B are attached to a wall with ...
- math - A beam having a length of 20 metres is pivoted at its mid point. A 200 ...
- physics HELPP!!!!!!! - A beam resting on two pivots has a length of L = 9.00 m ...
- physics - A steel beam (with linear mass density of 6.88 kg/m) has a total ...
- Equilibrium - A beam having a length of 20 metres is pivoted at its mid point. A...