Physics help on conditions for Static Equilibrium
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Physics help on conditions for Static Equilibrium

[From: ] [author: ] [Date: 11-06-14] [Hit: ]
the center of mass of the scaffold pulls 196N downward and is 2m away, meaning a 2m x 196N = 392 Nm clockwise torque.The first painter is 3m away and weights 735 newtons, meaning a 3m x 735N = 2205 Nm torque clockwise.The right rope pulls upward at 1100 Newtons and is 4 meters away so that is 4400 Nm COUNTERclockwise.The final source of torque is the second painter.......



2.

Again, since the rotation rate of the scaffold does not change, the net torque around any point along the scaffold must be zero. I'll choose the left rope as the point to sum torques to zero and clockwise to again be positive.

From the perspective of the left end, the center of mass of the scaffold pulls 196N downward and is 2m away, meaning a 2m x 196N = 392 Nm clockwise torque. The first painter is 3m away and weights 735 newtons, meaning a 3m x 735N = 2205 Nm torque clockwise. The right rope pulls upward at 1100 Newtons and is 4 meters away so that is 4400 Nm COUNTERclockwise. The final source of torque is the second painter. Who is standing X meters away and weights 735N for a 735X Nm torque clockwise.

Summing all the torques to zero (clockwise is positive):

392 Nm + 2205 Nm - 4400 Nm + 735X N = 0

-1803 Nm + 735X N = 0

X = 1803Nm / 735 N

= 2.45m

This is the distance from the left end. So his distance from the right end is:

4.00m - 2.45m = 1.55m
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