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Let's say I push with an angle of 45 degrees at an object with a force of 100 N. Since the angle is 45, we would have a torque $\tau = 100 * sin(135) * r = 71N*r$

Since the force was not perpendicular we will also have a linear acceleration. How big is the force responsible for that? Is it still 100N even though a part of it created a torque?

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  • $\begingroup$ The perpendicular part is a sine, the other part is obviously a cosine (of the angle made with the position vector). $\endgroup$ – AV23 May 9 '15 at 10:43
  • $\begingroup$ @AV23 Obviously, but that's not my question. If we would be able to lock the object from rotating, would it just move in one direcion? $\endgroup$ – lawls May 9 '15 at 10:49
  • $\begingroup$ The radial direction, yes. $\endgroup$ – AV23 May 9 '15 at 10:51
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The force doesn't get "used up" by creating a torque. The torque and the force exist simultaneously.

You correctly computed the torque due to that force. If that is the only force on this object, then the linear acceleration of the object is given by Newton's 2nd law: $$a = \frac{100~\rm N}{m}$$ where $m$ is the mass of the object. That is the linear acceleration regardless of how much torque that force also produces.

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