What Is the Minimum Force Required to Move a Crate with Static Friction?

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Homework Help Overview

The problem involves determining the minimum force required to move a crate on a horizontal surface, taking into account static friction. The crate's weight and the coefficient of static friction are given, along with an angle at which the force is applied.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for the applied force P and its dependence on the angle θ. There are attempts to find the minimum value of P, with some suggesting that θ should be zero for minimum force. Questions arise about the weight of the crate and its absence in the problem statement.

Discussion Status

The discussion is ongoing, with participants exploring different methods to find the minimum value of P, including differentiation and rearranging the equation. Some guidance has been offered regarding the use of calculus to find the derivative and maximize the denominator of the rearranged equation.

Contextual Notes

There is a noted lack of information regarding the weight of the crate, which may impact the ability to calculate specific values. The problem also implies constraints related to the angle of application and the conditions under which motion becomes impossible.

McDonell
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Homework Statement



A crate of weight Fg is pushed by a force P on a horizontal floor. (a) If the coefficient of static friction is μ s and P is directed at angle θ below the horizontal, show that the minimum value of P that will move the
crate is given by

P = usFgSecθ / (1 - usTanθ)

(b) Find the minimum value of P that can produce motion when μ s = 0.400,
If the angle were 68.2° or more, the expression for P would go to infinity and motion would become impossible.


The Attempt at a Solution



I was able to figure out how to get to P, but I cannot figure out how to find the minimum value of P. I am assuming that if they want the minimum value of P, theta would be equal to 0, since all of the force would be put along the horizontal. I am not sure where exactly to go from there though.
 
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Mind showing us what you got as P?
 
Pranav-Arora said:
Mind showing us what you got as P?

well P is the same thing as in my original post.

P = usFgSecθ / (1 - usTanθ)

Now I don't know how to find the minimum value of P
 
Do they provide the weight of the object?
 
LawrenceC said:
Do they provide the weight of the object?

No they do not
 
What is the derivative of P with respect to θ ?
 
McDonell said:
well P is the same thing as in my original post.

P = usFgSecθ / (1 - usTanθ)

Now I don't know how to find the minimum value of P

Woops, sorry, i must have missed it. (I was feeling sleepy when i posted my reply, sorry)[/size]

Finding minimum value requires the knowledge of Calculus, do you know how to find the derivative?

You can do it in an another way too. Rearrange the equation, write sec and tan in terms of sin and cos, you get:
P=\frac{μ_sF_g}{\cos(\theta)-μ_s\sin(\theta)}
For P to be minimum, the denominator should be maximum, so simply differentiate \cos(\theta)-μ_s\sin(\theta) with respect to θ.
 
Last edited:
Find equation of applied force equal to frictional force.
You will get an equation that is equal to a constant.
For minimum value of P, the other factor must be maximum.
 
Last edited:

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