A truck with a heavy load has a total mass of 7500 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2.
A.) What was the mass of the load? Ignore rolling friction.
How would I solve this problem?
Anonymous
Favorite Answer
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
electron1
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
lunchtime_browser
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