# Find load weight of the truck

## 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.

ƩM = 0
ƩF = 0

## The Attempt at a Solution

Firstly, I found the moment about point B.

ƩMB = -N(110 in) + (3600 lb)(65 in) - (WL)(21 in)

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

ƩFy = N + N -3600lb -WL

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

2N = 3600lb + WL
N = 1800lb + WL/2

I then plugged the value of N into the moment equation

ƩMB = -(1800lb + WL/2) + (3600lb)(65 in) - (WL)(21 in)

Solving for WL:

0 = -198000 lbft - 55 WL + 234000 lbft - 21WL
76WL = 36000 lbft

WL = 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

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I went over my calculations and found my mistake. I confused the force of the truck (3160lb) with the example question they gave me in which the truck had a force of 3600lb. Thus my calculations are as follows:

ƩMB = 0
0 = -N(110 in) + (310lb)(65 in) - WL(21 in)

ƩFy = 0
ƩFy = N + N - 3160lb - WL
2N = 3160lb + WL
N = 1580lb + WL/2

ƩMB = -(1580lb + WL/2)(110 in) + 205400lb - 21WL
= -173800lb in - 55WL = 205400 lb in - 21WL

55WL + 21WL = 31600 lb in

Wl = 31600 lb in/ (55 + 21)

WL = 415.789 lb