A 4.99 m long rod of negligible weight is attached on one end to a ball joint which allows the rod to rotate in all directions. The free end of the rod points 26◦ above the eastward horizontal direction. A 62.5 N force directed vertically Down is applied at the rod’s free end. What is the direction of the torque due to this force (relative to the pivot end of the rod)?

The answer is 91.18 Nm

Explanation:

Solution

Recall that

The length of the rod = 4.99 m

∅ = 26°

Force = 62.5N

Now,

T = r * F

The direction of the torque will be in horizontally northward

The torque magnitude is  T =r F sin θ

where ∅ will be the angle between r + F  θ= 163°

Therefore,

T = 4.99 * 62.5 * sin 163

T =91.18 Nm

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