How Does Moving a Boat Affect Scale Readings and Torque?

  • Thread starter Thread starter canicon25
  • Start date Start date
  • Tags Tags
    Scale Torque
AI Thread Summary
Moving a boat on a trailer affects scale readings and torque due to the change in weight distribution. Initially, the scale reads 48kg, but after moving the boat 0.15m back, it reads 37kg, indicating a shift in the center of mass. The torque equations set up for equilibrium show that the moments created by the forces have opposite signs, as one force causes clockwise rotation while the other causes counter-clockwise rotation. The scale's reading reflects the net force acting on the trailer but does not directly contribute to the torque calculation. Understanding these dynamics is crucial for accurately determining the mass of the boat and the forces at play.
canicon25
Messages
25
Reaction score
0

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?

boat.png






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.


 
Physics news on Phys.org
canicon25 said:

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?

boat.png

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?
 
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?
 
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:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top