# An object with a mass of 5 kg is hanging from an axle with a radius of 18 m. If the wheel attached to the axle has a radius of 8 m, how much force must be applied to the wheel to keep the object from falling?

Nov 1, 2017

The $\text{Force is about } 110 N$.

#### Explanation:

There is something wrong with the data in your question. You say the axle has a radius of 18 meters. So that means a diameter of 36 m. That would be about the length of 9 small cars (VW Beetles). And then the wheel is less, 8 m - only 2 small cars. I will work the problem with the data you provided.

Keeping it from falling requires establishing equilibrium. Equilibrium requires that the sum of torques equal zero. Another way to say that is that torques must have equal magnitude but opposite direction.

Clockwise torque = Counter-clockwise torque

The hanging mass has weight of $5 k g \cdot 9.8 \frac{m}{s} ^ 2 = 48 N$. This creates torque through its 18 m lever arm according to:
$49 N \cdot 18 m =$882 N*m

The force to be applied to the wheel creates torque through its 8 m lever arm according to:
$F \cdot 8 m$

So, using Clockwise torque = Counter-clockwise torque
$F \cdot 6 m = 882 N \cdot m$
$F = \frac{882 N \cdot m}{8 m} = 110.25 N$ or about 110 N

That is the weight of $\frac{110 N}{9.8 \frac{m}{s} ^ 2} = 11.2 k g .$

I hope this helps,
Steve