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You have two possible configurations to make push-ups. In both configurations our arms are placed at 90 degrees from our body. The first position is normal push-ups where. your hands are placed on the ground. The second push-up. position, you have your legs on the ground and your hands on a cube. Which one between those two configurations is the easiest and why according to force diagrams?

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I thought that maybe the first one is the easiest because your hands are directly on the ground and therefore only force you have to fight against is gravity. While in the second position your hands are place on a block and you have to maintain that block in position while doing push up, therefore you have more force to fight against. Maybe im wrong...

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Second one is definitely easy.Imagine your body like an inclined plane and theta is the angle made by your feet is theta with the floor. In second case your hand push up the cos component of your weight , while in first case your hand push up against $\approx$ mg.

You can verify this , when you stand straight , theta made by feet is $90^o$. and when you push yourself against the wall becomes relatively easy.

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When the person starts to move, all its weight must be supported by feet and hand:

The hands must do a force so that the torque be at least zero, otherwise the person falls down.

Let's $\theta$ be the initial angle between the person and the ground, W his weight (supposed acting half its height $h$). The force $F$ from the arms is directed perpendicular to the person as indicates the picture. The torque calculated from the feet is:

$$T = W\frac{h}{2}cos(\theta) - Fh$$

The first term in the right side decreases as $\theta$ increases. What means that less force is necessary to get zero torque. Fig 2 is easier.

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