- #1

Northbysouth

- 249

- 2

## Homework Statement

The indicated location of the center of gravity of the 3160-lb pickup truck is for the unladen condition. If a load whose center of gravity is x = 21 in. behind the rear axle is added to the truck, determine the load weight WL for which the normal forces under the front and rear wheels are equal.

I have attached an image of the question.

## Homework Equations

ƩM = 0

ƩF = 0

## The Attempt at a Solution

Firstly, I found the moment about point B.

ƩM

_{B}= -N(110 in) + (3600 lb)(65 in) - (W

_{L})(21 in)

Then I found the forces acting in the y-direction:

ƩF

_{y}= N + N -3600lb -W

_{L}

I then solved this equaiton (force equation) for N, giving me:

2N = 3600lb + W

_{L}

N = 1800lb + W

_{L}/2

I then plugged the value of N into the moment equation

ƩM

_{B}= -(1800lb + W

_{L}/2) + (3600lb)(65 in) - (W

_{L})(21 in)

Solving for W

_{L}:

0 = -198000 lbft - 55 W

_{L}+ 234000 lbft - 21W

_{L}

76W

_{L}= 36000 lbft

W

_{L}= 473.684 lb

But it says my answer is wrong. Even more annoying is that this online homework has an example question that is identical except for the numbers, and I ran through that doing the same calculations (with the numbers altered appropriately) and I got the same answer that they did.

Any suggestions or helpful hints would be appreciated. Thanks