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How to find the angle that will pull the most weight

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  • #1
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Homework Statement



Straight out of the book: "An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1100N. The coefficient of static friction between the box and the floor is 0.35. (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand and (b) what is the weight of the sand and box in that situation?"


Homework Equations



F = ma
Fs = μsFN

The Attempt at a Solution



So, I've decided that the x component, that is, the 1100N (I call this Fc) times cosθ, has to be equal to static friction. If it wasn't, then more sand could be added until it was (is this an incorrect assumption?). From there I tried to solve it like my professor has been doing by analzying the sum of the forces in the x direction and y direction separately to try to get two equations with two variables. The idea then would be that it would be an algebra problem, but I keep getting three unknowns (FN, Fg, and θ). I'll go ahead and show my work, I am probably going something wrong or going about it the wrong way (if I wasn't, this problem would have been done long ago).

Ʃ Fx = FcCosθ - Fs = 0

Or

FcCosθ = μsFN


Ʃ Fy = FcSinθ + FN - Fg= 0

or

FcSinθ + FN = Fg

So, I don't know θ, FN, or Fg and I don't see any third equation that I can use to eliminate a variable. Do I have to set the y component of Fc to something like I did the x component? Or was that my mistake from the beginning? Also, I will probably be spamming this forum with physics problems, sorry and if there is a limit do say. But I am tired of struggling with these homework problems, and the professor's hours are kind of awkward for me.
 

Answers and Replies

  • #2
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I think you are doing OK so far.
Use the first equation to replace F_N in the second. You'll have Fg, or the weight of the box, as a function of the angle (Fc is a constant).
Find the angle which maximizes that function.
 
  • #3
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FCosθ=Nμ .....(1)
N=mg-FSinθ .....(2)

Sub.(2) in (1).
FCosθ=(mg-FSinθ)μ
FCosθ=mgμ-FSinθμ

mgμ=F(Cosθ+Sinθμ)

Maximum value of (Cosθ+Sinθμ) is when Cosθμ-Sinθ=0
Edit: Errors . Thanks Nasu
 
Last edited:
  • #4
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418
You forgot to multiply N by Fsinθ when you multiplied the parenthesis (fourth equation).
 

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