How Does Moving a Boat Affect Scale Readings and Torque?

[canicon25]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

Taking moments about the wheel:

(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.

 

[vela]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

Taking moments about the wheel:

(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.

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.

Other than that, your set-up looks fine. What's your question?

 

[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?

 

[vela]

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.

 

[asteriatic]

To solve this problem, we first need to understand the concept of torque. Torque is a measure of the rotational force applied to an object, and it is defined as the product of the force and the distance from the pivot point. In this case, the pivot point is the wheel of the trailer, and the force is the weight of the boat and the trailer.

We can use the equation torque = F(perpendicular) * d to calculate the torque exerted by the boat and the trailer on the scale. Since the scale reads 48kg when the boat is in its original position, we can say that the torque exerted by the boat and the trailer is (48*9.81*6), where 6 is the distance from the pivot point to the scale.

When the boat is moved 0.15m further back on the trailer, the torque exerted by the boat and the trailer changes. The scale now reads 37kg, so the torque exerted is (37*9.81*6). We can set these two torques equal to each other and solve for the mass of the boat (m).

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

Solving for m, we get m = 11.5kg. Therefore, the mass of the boat is 11.5kg.

In order to check our answer, we can also use the equations for equilibrium: Sum moments = 0, Sum Fy = 0, and Sum Fx = 0. These equations state that in order for an object to be in equilibrium, the sum of the moments, forces in the y-direction, and forces in the x-direction must all equal zero.

Using these equations, we can solve for the unknowns (m and x) and check if our answer is correct. The calculations will show that our answer of 11.5kg for the mass of the boat is indeed correct.