# Truck problem

1. Jan 11, 2004

### tandoorichicken

A trucker needs to weigh a truck that is too long to fit on a platform scale. When the front wheels of the truck are run onto the scale, the scale reads $$W_1$$. When the rear wheels are run onto the scale so that the front wheels are off, it reads $$W_2$$.
a) prove that the total wight of the truck is $$W_1 + W_2$$.
b) prove that if the truck is loaded so that its center of gravity is halfway between the front and rear wheels, the total weight is $$2W_1$$.

2. Jan 12, 2004

### ShawnD

Here is my FBD for case 1, the scale is on the left.

force up (W1)------x distance-------centre of grav (G)---d distance----wheels (A)

The moment around A looks like this

$$0 = dG - W_1(d + x)$$ no movement

$$W_1(d + x) = dG$$ isolating for W1

$$W_1 = \frac{dG}{d + x}$$ isolating for W_1

Now here is case 2 where the scale is the on the right.

wheels (A)-----x distance-------centre of grav (G)---d distance-----force up (W2)

Moment around the new A

$$0 = W_2(d + x) - xG$$ no movement

$$W_2(d + x) = xG$$ isolating W_2

$$W_2 = \frac{xG}{d + x}$$ isolating W_2

Now you are supposed to prove that W1 + W2 = G so just fill in now

$$W_1 + W_2 = G$$ simple version

$$\frac{dG}{d + x} + \frac{xG}{d + x} = G$$ expanded version

$$\frac{dG + xG}{d + x} = \frac{G(d + x)}{d + x}$$ common denominator

$$\frac{dG + xG}{d + x} = \frac{dG + xG}{d + x}$$ expanded

They are the same are they not? :D

Once you solve A, B is very easy. If G is to equal 2W1, all you need to prove is that W2 and W1 are equal. If the centre of gravity is in the middle, the equations for W1 and W2 will be the exact same.

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