Calculating the Upward Force on Support B in a Bridge-Torque Problem

[mburt]

Homework Statement

The diagram below shows a 2.0 × 10³ kg truck on a 20.0 m long uniform bridge that has a mass of 8.0 × 10³ kg. If the truck is 6.0 m from support A, what is the magnitude of the upward force at support B?

[PLAIN]http://img534.imageshack.us/img534/6554/torquev.png

Homework Equations

T = rFsin(x)

The Attempt at a Solution

[PLAIN]http://img17.imageshack.us/img17/5487/torqueattempt.png

T_clockwise = T_{counterclockwise}
T_truck + T_bridge = T_B
(6.0m)(2000 kg)(9.80 m/s^2) + (10.0m)(8 000kg)(9.80m/s^2) = (20.0m)F
90160 Nm / 20.0m = F
F = 4.5x10⁴ N

However the answer given is 4.6x10³ N.

Any thoughts?

 

[WatermelonPig]

sum(torque) = 0

(6(mass-truck) + 10(mass-bridge))*g - 20FB = 0

FB = (9.8/20))*(6*2000+10*8000) which is about what you got so I don't know maybe they are using g = 9.81 and then are off by a factor of 10.

 

[mburt]

Maybe it's an incorrect answer... Hmm I'll wait for some other people's input too.

Thanks!

 

[Quinzio]

There is a mistake in the book answer. If you multiply the book answer by 9.8 you get your answer.

 

[ashishsinghal]

Your method is correct.