How Do You Calculate Load Weight for Balanced Normal Forces in a Truck?

[Northbysouth]

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

 

[Northbysouth]

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:

ƩM_B = 0
0 = -N(110 in) + (310lb)(65 in) - W_L(21 in)

ƩF_y = 0
ƩF_y = N + N - 3160lb - W_L
2N = 3160lb + W_L
N = 1580lb + W_L/2

ƩM_B = -(1580lb + W_L/2)(110 in) + 205400lb - 21W_L
= -173800lb in - 55W_L = 205400 lb in - 21W_L

55W_L + 21W_L = 31600 lb in

W_l = 31600 lb in/ (55 + 21)

W_L = 415.789 lb

This is the correct answer