Find mass of load that falls off truck

[Andy Salter]

Homework Statement

A truck with a heavy load has a total mass of 7500 kg. It is climbing a 15 degree 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/s 2 . What was the mass of the load? Ignore rolling friction.

Homework Equations

ΣF = Fg + n + F1 = ma , where n is the normal force and F1 is the parallel force

The Attempt at a Solution

Before the load falls off we know a = 0 therefore
ΣF = Fg + n + F1 = 0

Fg before fall = 9.81 m/s^2 * 7500 kg = 73575 N

Fg_x before fall = -73575*Sin(15) = -19042

Because ΣFx = 0 before fall, Fg_x = -F1_x

=> F1_x = 19042 N

After fall we have:

Fx = 19042 + Fgx = FgSin(15) = m*1.5
Fy = 0

I'm not sure where I need to go from here. I need to find m so I can find the difference, but I can't do this without either the magnitude of Fg or Fgx. Not sure how I can find these...

 

[TSny]

Welcome to PF!

Is there a way to express F_g in terms of m?

 

[Andy Salter]

Ok so I assume you mean F_g = ma = 9.18m

I guess this gives me these simultaneous equations:

Fg = 9.81m
Fg = 1.5m / Sin(15)

But the solution of this is Fg = m = 0, so there's something wrong

 

[TSny]

Fg = 9.81m

OK

Fg = 1.5m / Sin(15)

How did you arrive at this?

 

[Andy Salter]

Fx = 19042 + Fgx = FgSin(15)
Fy = 0
=> FgSin(15) = Fx = F = ma = 1.5m
=> Fg = 1.5m/Sin(15)

 

[TSny]

Fx = 19042 + Fgx = FgSin(15)

The first equality looks good. But I don't understand the second equality. How did you get 19042 + Fgx = FgSin(15)?

 

[Andy Salter]

My understanding was that F_x = Fg_x = FgSin(15)

 

[Andy Salter]

Oh, wait, I see why that is wrong haha

 

[Andy Salter]

F_x = 19042 + Fg_x = 19042 + FgSin(15) = 1.5m

which gives the simultaneous equations:

Fg = 9.81m
Fg = (1.5m - 19042)/ Sin(15)

With solutions:

f = -46249 and m = 4714.52

Therefore:
Δm = 7500 - 4714.52 = 2785.48 kg

Yes?

 

[TSny]

F_x = 19042 + Fg_x = 19042 + FgSin(15) = 1.5m

Does Fg_x = +FgSin(15) or does Fg_x = - FgSin(15)?

which gives the simultaneous equations:Fg = 9.81m
Fg = (1.5m - 19042)/ Sin(15)

With solutions:

f = -46249 and m = 4714.52

Therefore:
Δm = 7500 - 4714.52 = 2785.48 kg

Yes?

Basically this is correct except for sign errors stemming from the wrong sign for Fg_x. I think your results for m and Δm are correct.

 

[Andy Salter]

Ah yes, you are indeed right. Thankyou for your help :)

 

[TSny]

OK. Good work.

When solving the two equations

Fg = 9.81m
19042 - FgSin(15) = 1.5m

it might be easier to just substitute the expression for Fg from the first equation into the second equation. So, the second equation becomes

19042 - 9.81m Sin(15) = 1.5m

Then you can solve this for m. But your method of doing the algebra is OK.