Inclined Plane and box

  • Thread starter jerryez
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  • #1
20
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The owner of a warehouse ask an engineer to design a ramp which will reduce the force necessary to lift boxes to the top of a 1/2 m step. If there is only room enough for a 4 m ramp, what is the maximum factor by which the lifting force could be reduced?




I am not sure how to find the maximum factor?? Could anyone help?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
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hi jerryez! :wink:

well, the force to lift it without a ramp, vertically upwards (and with zero acceleration), is mgh …

what is the force needed if you use a 4m ramp? :smile:
 
  • #3
20
0
Hello Tiny-tim

Would the force than be equal to mgsintheta?? I dont think this is right but that is my best guess.
 
  • #4
656
2
You got it...

Now find theta and compare g to g*sintheta
 
  • #5
20
0
So theta = 7.2

g = 9.81 m/s(squared)

g*sintheta = 1.23


I still dont understand what the maximum factor by which the lifting force could be reduced is??
 
  • #6
rl.bhat
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4,433
7
It is a simple machine.

The mechanical advantage = Load/ effort = Distance traveled by the effort/ Distance traveled by the load. = 4/(1/2)
 
  • #7
tiny-tim
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hello jerryez! :smile:

(just got up :zzz: …)

The reason for the word "maximum" is because it depends on the angle of the force (and also on the acceleration being zero).

Usually when you push or pull something, your applied force isn't exactly along the slope, and if it's not exact, then you need extra force. :wink:
jerryez said:
Would the force than be equal to mgsintheta?? I dont think this is right but that is my best guess.
hmm … guessing will be no good in the exam :redface:

the method to use, as nearly always, is good ol' Newton's second law …

we're assuming no acceleration along the slope

so (let's call that direction "k") (Ftotal).k = ma.k = 0. :wink:
 
Last edited:
  • #8
656
2
Along with the other things Tim mentioned, maximum would also assume there is no friction. This would make a signficant difference in the real world. So I assume your problem also avoids this variable as well as if you did not pull the mass parallel to the plane or accelerate the mass up the plane.

I am sorry for the intervention. The posters question went unanswered for a while. So I just added that g v. sintheta*g gives you the appropriate ratio. If its force it would be mg v. m*sintheta*g, but since m is on both sides I took an irresponsible shortcut. Sorry.
 
  • #9
656
2
It is a simple machine.

The mechanical advantage = Load/ effort = Distance traveled by the effort/ Distance traveled by the load. = 4/(1/2)
I think its says the forced reduced by, so using your shortcut I think it would be 1/2 over 4. If it said how much "easier"...
 
  • #10
20
0
Thank you everyone for your help!! The answer to the problem is 8 I just didnt understand how they got it.

It is a simple machine.

The mechanical advantage = Load/ effort = Distance traveled by the effort/ Distance traveled by the load. = 4/(1/2)
I also see now how the ratio between g and g*sintheta also results in the correct answer.

9.81/1.23 = 7.9
 

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