1. Oct 15, 2016

### Rob123456789

1. The problem statement, all variables and given/known data
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.

3. 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.

2. Oct 15, 2016

### Rob123456789

Why is this wrong ?

3. Oct 15, 2016

### Staff: Mentor

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:

$$a = \frac{F_0}{5100 - M_L}$$

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?

4. Oct 15, 2016

### Rob123456789

F = (5100 – m) * 1.5 = 7650 – 1.5 * m so here is were I messed up ?

5. Oct 15, 2016

### Staff: Mentor

I think it might help to write the two sets of equations, for before and after the load has fallen off. Write out

$$∑F = ma$$

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