- #1
buffgilville
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There is a 0.8m long, 1.2 iron bar with a string attached to the center, making a 35 degree angle (from the bar). Calculate the torque produced by the tension in the string. A pivot at one end holds the bar, and the string is attached at the bar's center, which is the center of mass.
I'm confused because my book gave me two formulas for torque. Which one is right for this problem?
T=F*s*sin
= (1.2 * 9.81) (0.8) (sin35)
= 5.402 N*m
T=Fr
= (1.2*9.81)(0.4)
= 4.7088 N*m
I'm confused because my book gave me two formulas for torque. Which one is right for this problem?
T=F*s*sin
= (1.2 * 9.81) (0.8) (sin35)
= 5.402 N*m
T=Fr
= (1.2*9.81)(0.4)
= 4.7088 N*m