Help: TORQUE

buffgilville

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

Parth Dave

Well both formulas are infact identical. In the second one, you were probably missing the little perpendicular to sign as a subscript. That indicates that you only multiply the perpendicular component. Which is exactly what the first formula does (the sin35 will yield only the perpendicular component). With that being said, in this situation it would be better to use the first formula (since the moment arm and the force are not perpendicular). However, you seem to have made a small mistake. Remember, the moment arm is the distance in between the rotation axis and the force vector.