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