Mass of load falling from truck on 15° incline

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Homework Statement


A truck with a heavy load has a total mass of 5100 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.

What was the mass of the load? Ignore rolling friction.
Express your answer with the appropriate units.

The Attempt at a Solution


Force parallel = 5100 * 9.8 * sin 15 = 12935.78 N
Let m be mass of load. The final mass of the truck is 5100 – m.
Since the truck is accelerating at 1.5 m/s^2, let’s multiply this by 1.5.

F = (5100 – m) * 1.5 = 7650 – 1.5 * m
Set this equal to 12935.78 and solve for m.
7650 – 1.5 * m = 12935.78
1.5 * m = 5285.78
m = 5285.78 ÷ 1.5
This is approximately 3523.853 kg.
 
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Rob123456789 said:

Homework Statement


A truck with a heavy load has a total mass of 5100 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.

What was the mass of the load? Ignore rolling friction.
Express your answer with the appropriate units.

The Attempt at a Solution


Force parallel = 5100 * 9.8 * sin 15 = 12935.78 N
Let m be mass of load. The final mass of the truck is 5100 – m.
Since the truck is accelerating at 1.5 m/s^2, let’s multiply this by 1.5.

F = (5100 – m) * 1.5 = 7650 – 1.5 * m
Set this equal to 12935.78 and solve for m.
7650 – 1.5 * m = 12935.78
1.5 * m = 5285.78
m = 5285.78 ÷ 1.5
This is approximately 3523.853 kg.
Rob123456789 said:
Why is this wrong ?
I think there is a term missing. Think about the situation. If the mass of the load M_L were zero, you should get zero acceleration for the equation relating net acceleration to M_L. If you write the equation for acceleration a as a function of M_L given the way you have laid it out so far, you would get something like this:

[tex]a = \frac{F_0}{5100 - M_L}[/tex]

You can see that as M_L goes to zero in this equation, you still have an acceleration. What term might be missing from this equation that would make a-->0 as M_L -->0? Does that help you re-write your equations?
 
F = (5100 – m) * 1.5 = 7650 – 1.5 * m so here is were I messed up ?
 
I think it might help to write the two sets of equations, for before and after the load has fallen off. Write out

[tex]∑F = ma[/tex]

for before and after. Be sure to include all forces acting on the truck... :smile: