Find mass of load that falls off truck

AI Thread Summary
A truck with a total mass of 7500 kg accelerates at 1.5 m/s² after its load falls off while climbing a 15-degree incline. The forces acting on the truck before and after the load falls are analyzed using equations of motion, leading to simultaneous equations for gravitational force (Fg) and the mass of the load (m). The calculations reveal that the mass of the load is approximately 2785.48 kg, after correcting for sign errors in the force equations. The discussion emphasizes the importance of accurately applying physics principles to solve for unknowns in dynamics problems. The final mass of the load is confirmed through algebraic manipulation of the equations.
Andy Salter
Messages
17
Reaction score
1

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

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

Because ΣFx = 0 before fall, Fgx = -F1x

=> F1x = 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...
 
Physics news on Phys.org
Welcome to PF!

Is there a way to express Fg in terms of m?
 
Ok so I assume you mean Fg = 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 o_O
 
Andy Salter said:
Fg = 9.81m
OK

Fg = 1.5m / Sin(15)
How did you arrive at this?
 
Fx = 19042 + Fgx = FgSin(15)
Fy = 0
=> FgSin(15) = Fx = F = ma = 1.5m
=> Fg = 1.5m/Sin(15)
 
Andy Salter said:
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)?
 
My understanding was that Fx = Fgx = FgSin(15)
 
Oh, wait, I see why that is wrong haha
 
Fx = 19042 + Fgx = 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?
 
  • #10
Andy Salter said:
Fx = 19042 + Fgx = 19042 + FgSin(15) = 1.5m
Does Fgx = +FgSin(15) or does Fgx = - 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 Fgx. I think your results for m and Δm are correct.
 
  • #11
Ah yes, you are indeed right. Thankyou for your help :)
 
  • #12
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.
 
Back
Top