Atrailer carrying a boat is supported by a scale which initially reads 48 kg. the boat (and therefore its centre of gravity) is moved 0.15 m further back on the trailer. the scale now reads 37 kg. find the mass of the boat.

a. 3400 kg b. 1700 kg c. 1600 kg d. 440 kg

440 kg
view the diagram of the torques

let fg be the force of gravity exerted by the boat from its center of gravity
let x be the initial distance from the wheel pivot point.

rotational equilibrium so net torques ∑τ are equal to zero

we choose the pivot point to be the wheel so that we take advantage of both the 6m and the scale readings.

the scale readings mean that much mass is on it, so that means there is some gravity on it: mg = m(9.81m/s²)

before the boat is moved
∑τ = fg*x + (48 kg)(9.81 m/s²)(6 m) = 0 (i)

after the boat is moved
∑τ = fg*(x-0.15 m) + (37kg)(9.81 m/s²)(6m) = 0 (ii)

solve for x in (i)

fg * x + (48 kg)(9.81 m/s²)(6 m) = 0
fg * x  + 2825.28 nm = 0
x = -2825.28 nm / fg

substitute into (ii)

fg*(x-0.15m) + (37kg)(9.81 m/s²)(6m) = 0
fg*[ (-2825.28 nm / fg) - 0.15m] + 2177.82 nm = 0
-2825.28 nm - fg 0.15 m + 2177.82 nm = 0

-fg 0.15 m = 647.46 nm
fg = 4316 . 4 n

so the mass is

fg = mg
m = fg / g
m = (4316.4 n) / (9.81 m/s²)
m = 440 kg

440 kg

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