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Imagine a car with a crane.

In scenario 1, the tension on the crane cable by the load points straight down, ie, perpendicular to the ground.

In scenario 2, the tension on the crane cable by the load points 30 degrees outward from a line perpendicular to the ground.

Mathematically, it is easy to calculate that the crane can support a much bigger load without the car tipping in scenario 1. But what is the explanation of this? May I explain it like this: The tension gains an x component in the second scenario, thus some of the weight of the car must compensate for this. Or is there an easier way to explain this?

Akitirija
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1 Answers1

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The torque is the product of the force and the perpendicular distance from the force to the support. When you pull at an angle, you are in effect increasing the "lever" with which you are rotating. Just as you can tighten a nut better with a long handled wrench than with a short one.

This diagram may help:

enter image description here

The magnitude of $F_1$ and $F_2$ is the same, but $F_2$ is pulling at an angle. As you can see, although this makes the vertical component of the force a bit smaller ($F_v = F_2 \cos\theta$), you get a larger torque because of the increased lever ($d$ becomes $d'=d+h\sin\theta$).

The greatest torque (about the wheel W) is applied when we maximize

$$\Gamma = d' F_2 \cos\theta = (d + h\sin\theta)F_2 \cos\theta$$

Obviously, when $h=0$ the torque will be greatest when $\cos\theta = 1$; but as $h$ gets bigger, the angle at which you can pull and get a greater torque increases.

Floris
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