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A uniform metal rod of mass 100 kg and length L (=2.5 meters) is suspended from the side of a building as shown in the figure. On the far end a mass of 100 kg is hung by a rope, (distance L from the building). The rod is connected to the building on the left end by a pivot, (and a force F is extend on the rod by the pivot). To hold thips up a strong cable is attached a distance L/4 from the pivot (and has a tension T on it).
a)The system is in equilibrium.
Using rotational equilibrium and torque, find the tension in the cable.
b) Using net forces, calculate the force the pivot exerts on the rod.

2007-07-30 06:43:53 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

Since there is no figure included I will assume that the rod is horizontal

The angle between the cable and the rod is th, which is
th=tan-1(2.5/10)

Summing the torques at the pivot:
-T*cos(th)*2.5/4+
100*2.5/2+
100*2.5=0

Solve for T which is the tension in the cable

T=(600/cos(th))g

618*9.81N

b) The rod pivot has horizontal force equal to
-T*cos(th)
=600*9.81 N

and vertical force of
F-200+T*sin(th)=0
=50 *g N
j

2007-07-30 09:34:31 · answer #1 · answered by odu83 7 · 0 0

where is the figure?

2007-07-31 10:08:01 · answer #2 · answered by mramahmedmram 3 · 0 0

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