Physics Heavy Load Problem?

Mass of truck = M kg

Mass of load = m kg

Thrust from engine = T

Before the load falls off, thrust (T) balances thecomponent of total weight downhill:

T = (M+m)gsinθ (equation 1)

After the load falls off, thrust (T) remains the same but downhill component of weight becomes Mgsinθ .

Resultant force on truck is F = T – Mgsinθ

F causes the acceleration of the truck; using ‘force = mass x acceleration’ gives:

T – Mgsinθ = Ma

T = M(a + gsinθ) (equation 2)

Combining equations 1 and 2:

(M+m)gsinθ = M(a + gsinθ)

Mgsinθ + mgsinθ = Ma + Mgsinθ

mgsinθ = Ma (equation 3)

Since M+m = 7500kg, M = 7500 – m. Equation 3 becomes:

mgsinθ = (7500 – m)a

. . . . . . = 7500a – ma

mgsinθ + ma = 7500a

m(gsinθ + a) = 7500a

m = 7500a/(gsinθ + a)

It depends what value is used for g; I’ll use 9.8m/s².

m = 7500 x 1.5 / (9.8sin(15°) + 1.5)

. . = 2787kg

. . = 2800kg to 2 significant figures

For the truck to move up the incline at a steady speed, the force must be equal to the component of its weight that is parallel to the incline.

Force parallel = 7500 * 9.8 * sin 15 = 73,500 * sin 15

This is approximately 19,023 N. This force causes the unloaded truck to accelerate. Let m be the mass of the load.

73,500 * sin 15 = (7500 – m) * 1.5

1.5 * m = 73,500 * sin 15 – 11,250

m = (73,500 * sin 15 – 11,250) ÷ 1.5 = 49,000 * sin 15 – 7500

The mass of the load is approximately 5,182 kg

Call mass of load (M); mass of truck (m)

So

M + m = 7500 kg

weight of load + truck = (M+m)g = (7500 kg * 9.8 m/s²) = 73500 newtons

component of weight down slope = 73500.sin(15°) = 19000 N

" It is climbing a 15∘ incline at a steady 15 m/s.." So, the thing is not accelerating. Ignoring friction, the net force on the rig must be zero.

So the traction force up the hill = component of weight force down the hill = 19000 N

Now it sheds its load. The mass (m) is still under the same traction force but the component of the weight force down the hill is now mg.sin(15°) = 2.54m newtons

Net force accelerating truck = (19000 - 2.54m)

Newton's second law F = m a

19000 - 2.54m = m * 1.5

19000 = 4.04m

m = 4700 kg

M = 7500 - 4700 = 2800 kg {2 sig figs}

Fnet = ma = 0 = Ftruck - Fgrav

Ftruck = 7500(9.81)sin 15

F truck = 19,042.6 N

Fnet = ma = 1.5x = 19,042.6 - x(9.81)sin 15

4.04x = 19,042.6

x = 4,714.66 kg (truck)

load = 2,785.34 kg