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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)?

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Answer:

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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