 #1
cs44167
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 Homework Statement:
 A rigid beam of mass 7.78 kg and length 1.44 m is fixed at point P, around which the beam rotated (P is at the far leftmost point of the beam). What is torque on the beam due to gravity only?
 Relevant Equations:

Torque = r*F*sin (theta)
w = mg
I tried using r * f * sin theta and calculated this:
1.14 m * 9.80 m/s/s * 7.78 kg = 109.7 N*m
this was wrong; I needed three significant figures so I did 1.10E2 N*m which was also wrong.
Since the torque is due to gravity; would it be 1.10E2 N*m since it’s angle is 270° which is 1?
1.14 m * 9.80 m/s/s * 7.78 kg = 109.7 N*m
this was wrong; I needed three significant figures so I did 1.10E2 N*m which was also wrong.
Since the torque is due to gravity; would it be 1.10E2 N*m since it’s angle is 270° which is 1?