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My teacher told me that, by using moments, we can see how it takes half the force to lever an object up.

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  • $\begingroup$ Can you elaborate on this, maybe provide an example? As it is, your statement isn't quite true. $\endgroup$ – Jakob Lovern Dec 8 '17 at 18:51
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Torque

Recall that $\tau = F\cdot d$. The closer an object is to the fulcrum, the greater its effective torque. If you have a mass of 15 kg, it would take a force greater than 147.15 N (ish) to lift it vertically. Now, assume that the mass is on a lever. It is one meter from the fulcrum, ^. On the other side, you push down two meters from the fulcrum. See the ASCII diagram:

V             O
---------------
  (2M)   ^ (1M)

Now, the mass pushes down with a force of $a_g\cdot m = 9.81\cdot15 = 147.15$ N. The clockwise torque on the fulcrum, then, is $\tau = F\cdot d = 147.15\cdot1 = 147.15\ \mathrm{N\cdot m}$. To oppose this torque, there would need to be a counterclockwise torque of the same magnitude. Using the torque equation, we can see that when we double the distance, the magnitude of the force has to be cut in half to keep torque constant.

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Suppose you have a cube on a flat surface with bottom vertices A, B, C, and D on the surface. Suppose you apply force F directed upwards to the cube at point K in the middle of edge AB. You can consider equilibrium of the cube when the forces acting on the cube are F at point K (directed upwards), -mg at point O (center of mass of the cube) (directed downwards) and normal reaction N at point L in the middle of edge CD (directed upwards). If you calculate moments with respect to edge CD in equilibrium, you'll see that in equilibrium F=mg/2.

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The intuitive answer: Suppose we have Point A as the fulcrum (the point that the object is being rotated around), and Point B as the point the force is being applied. Then the mass at Point A isn't moving, and the mass at Point B is moving at normal speed. On average, the mass is moving at half speed of Point B, and the object moves, on average, half the distance that Point B does. Since energy is force times distance, the force at Point B times the distance that Point B travels is equal to the force on the object as a whole times the distance that the object as a whole travels. Since the object travels half the distance that Point B does, the force at Point B is half the force on the object.

As @ Jakob Lovern mentions, however, this statement requires further qualifications to be true, such as that the mass is homogeneously distributed, and it only applies at the very first instant. Once you start levering an object, your force starts to become partially horizontal, so the force needed decreases even further.

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