Calculate the average force used to drag a box up a ramp

AI Thread Summary
The discussion centers on calculating the average force required to drag a box up a ramp, with participants questioning the provided equations and variables. There is confusion regarding the definitions of force (F or Fa) and work (W), as well as the use of cos(20) in the calculations. Participants point out that the question may be flawed since it assumes constant force without sufficient information. Clarification is sought on the direction of the force and how the length is measured in relation to the ramp's angle. The conversation highlights the need for more precise definitions and information to solve the problem accurately.
RyanRhino
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Homework Statement
A 150. kg box is dragged up a 4.00 m long ramp inclined at 20.0 degrees to the horizontal. You are told that the efficiency of the ramp is 55.0%
a) find the average force that is used to drag the box up the ramp.
Relevant Equations
W = (FacosX)(d)
%Eff. = Eout/Ein x 100%
I first wrote down that 55% = Eout/Ein
I also know that W = (Facos20)(4)
and I substitute it into the first equation
55% = Eout/[(Facos20)(4)]
But I'm missing two variables here. Did i forget something or is the question missing some information?
 
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RyanRhino said:
W = (Facos20)(4)
How are you defining F there (or is it Fa?)? I ask because I am unsure how you arrive at cos.
Would that W be Ein or Eout?
What information have you not used?

Apart from all that, the question is wrong in asking for the average force. With the given information, you can only solve it by assuming the force is constant.
 
haruspex said:
How are you defining F there (or is it Fa?)? I ask because I am unsure how you arrive at cos.
Would that W be Ein or Eout?
What information have you not used?

Apart from all that, the question is wrong in asking for the average force. With the given information, you can only solve it by assuming the force is constant.
So F there i believe is Fa and that W would be the total work put in which would be Ein
 
RyanRhino said:
So F there i believe is Fa and that W would be the total work put in which would be Ein
You have not explained how you get the cos(20) in there.
What direction are you supposing the force to be in? In what direction is the length measured?

How can you calculate Eout? What was achieved by dragging the box up?
 
haruspex said:
You have not explained how you get the cos(20) in there.
What direction are you supposing the force to be in? In what direction is the length measured?

How can you calculate Eout? What was achieved by dragging the box up?
Sorry i forgot, the cos 20 is from the ramp of how it is 20 degrees above the horizontal we just pushed the box up the ramp
 
RyanRhino said:
Sorry i forgot, the cos 20 is from the ramp of how it is 20 degrees above the horizontal we just pushed the box up the ramp
That is not what I meant. I know the ramp is at 20 degrees, but what is your reasoning for including the cosine of it in that expression?

Try to answer my questions:
What direction are you supposing the force to be in?
In what direction is the length measured?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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