# Torque problem with scale

1. Dec 17, 2011

### canicon25

1. The problem statement, all variables and given/known data

A trailer carrying a boat is supported by a scale which initially reads 48kg. The boat is moved 0.15m further back on the trailer. The scale now reads 37kg. What is the mass of the boat?

2. Relevant equations

torque = F(perpendicular) * d
For equilibrium:
Sum moments = 0
Sum Fy=0
Sum Fx=0

3. The attempt at a solution

(48*9.81*6)+9.81*m*x=0
(37*9.81*6)+(9.81*m*(x-0.15))=0

Two equations and two unknowns (m and x). Solve.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 17, 2011

### vela

Staff Emeritus
In both cases, the two moments should have the opposite signs since one force wants to cause the trailer to rotate clockwise, and the other, counter-clockwise.

3. Dec 17, 2011

### canicon25

my original question was simple am i going about setting up the equation correctly.

but my question now is why are the two moments opposite in sign.

if there is a measure of 46kg on the scale why is it not a component of the force acting down on the scale?

4. Dec 17, 2011

### vela

Staff Emeritus
Because you're calculating the moment acting on the trailer. The trailer pushes down on the scale, but that force doesn't exert a torque on the trailer. It's the reaction force, the scale pushing up on the trailer, that is responsible for the torque on the trailer.

Last edited: Dec 17, 2011