What Angle Minimizes Force to Make a Block Slip?

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To determine the angle theta that minimizes the force required to make a block slip, the relationship between the applied force, friction, and normal force must be analyzed. The frictional force is given by Ff = μN, where N = Mg - Fsin(theta). By substituting this into the equation and expressing F as a function of theta, the derivative can be set to zero to find critical points. The solution reveals that tan(theta) = μ, indicating that the angle for minimum force is dependent on the coefficient of friction. A suggestion is made that the minimum angle could be zero, but this does not align with the derived relationship.
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Homework Statement


A person pulls on a block with a force F, at an angle theta with respect to the horizontal. The coefficient of friction between the block and the ground is mu. For what theta is the F required to make the block slip a minimum.


Homework Equations



Net force equations for static objects

The Attempt at a Solution



So I'm pretty sure I've got all of my equations right, but I'm having trouble simplifying my results further.
Ff=\mu N when the box slips and N=Mg-Fsin\theta Substitution gives \mu\left(Mg-Fsin\theta\right)=Fcos\theta I don't know how to solve for theta.
 
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For theta for minimum F, write F as an explicit function of theta and differentiate w.r.t theta. Put it equal to zero(the calculus concept of maxima and minima).
 
But I'm not trying to find the minimum F, I'm trying to find the minimum theta.
 
Ok I think I have it now. I maximize the denominator in F=\frac{\mu Mg}{cos\theta+\mu sin\theta} so the derivative is \mu cos\theta-sin\theta=0 So tan\theta=\mu
 
As an afterthought, don't you think the minimum theta is zero?
 
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