Calculating Torque of a String Attached to an Iron Bar

[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.

 

[wais]

Both formulas for torque are correct, but they are used in different situations. The first formula, T=F*s*sin, is used when the force is not applied perpendicular to the lever arm (in this case, the distance from the pivot to the point where the force is applied). The second formula, T=Fr, is used when the force is applied perpendicular to the lever arm. In this problem, since the string is attached at the center of mass of the bar, the force is applied perpendicular to the lever arm, so the second formula should be used. However, if the string was attached at a different angle, the first formula may need to be used. It is important to understand the differences between the two formulas and when to use them. In this case, using the second formula, the torque produced by the tension in the string is 4.7088 N*m.