Min/Max angle of a force applied to stationary mass.

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SUMMARY

The discussion focuses on determining the maximum and minimum angles (theta) at which a constant horizontal force (Fh) can be applied to a block of mass (m) against a vertical surface while keeping it stationary. Key equations involve the balance of forces in both the x and y directions, incorporating static friction (mu), gravitational force (mg), and the components of the applied force (Fh). The resolution of the equations leads to a requirement for calculus to find the extrema of theta, providing a clear path to solving the problem.

PREREQUISITES
  • Understanding of static friction and its coefficient (mu)
  • Knowledge of free-body diagrams and force resolution
  • Familiarity with Newton's laws of motion
  • Basic calculus for finding maxima and minima
NEXT STEPS
  • Study the principles of static friction and its role in equilibrium
  • Learn how to construct and analyze free-body diagrams for complex systems
  • Review Newton's laws of motion in the context of forces acting on stationary objects
  • Explore calculus techniques for finding extrema, particularly in physics applications
USEFUL FOR

This discussion is beneficial for physics students, engineers, and anyone involved in mechanics or force analysis, particularly those working with static systems and frictional forces.

pirland
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Ok, I've been having some real trouble with this problem. Either I’m missing some key piece of reasoning or it’s harder then it seems.

If a block of mass m is pressed against a vertical surface by a constant force Fh, what are the maximum and minimum angels that that force can be applied and still have the block remain stationary? Express in terms of m, g, theta, and mu.”

Now I assume that what is needed is to equate the maximum static friction force to the total force applied parallel to the surface, but I’m having trouble finding an equation that will give me my answer. Any tips or pointers would be greatly appreciated.
 
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Draw a free-body diagram for the block.

You have two components of force going down, gravity and the y component of the applied force. Gravity is given by mg and, supposing theta is the angle of elevation, the y-component of the applied force is F\sin{\theta}. The normal force is the only force going up, and is opposite and equal to the sum of the downward forces, since there cannot be any net y-component of force.

You have one force in each horizontal direction. On one side you have the x-component of applied force, F\cos{\theta}. On the other side you have the force of static friction, F_n \mu, where F_n is the normal force. Again, these two forces must balance.

Can you take it from there?

--J
 
Clarification

Sorry, I should have specified more, the problem that I'm having is with the resolving of the actual equation. Since the block has not passed through the wall the horizontal force is being countered by the normal force, causing balance in the x axis. What I believe I'm looking for is a way to solve for theta in a situation where the static force is exactly equal to the sum of all forces in the y axis. When I put that together mathmatically I get something along the lines of:

-cos(theta)Fh*mu=sin(theta)Fh+Fg

But I can't seem to resolve it into anything useful, mainly the maximum and minimum values of theta. If the above equation is in fact correct could you point me in the righ direction for solving it?
 
Oh, I totally misread the question. Sorry about that.

In the x-direction, there's the applied force and the normal force.

0 = F_h \cos{\theta} + F_N = 0

Note that the normal force will be negative, as it's pointing in the opposite direction.

In the y-direction, there's the applied force, gravity, and the static friction force.

0 = F_h \sin{\theta} + F_g + F_s

F_g, as I'm sure you know, is mg. F_s = F_N \mu, so you must solve the first equation for F_N and substitute it in there. This gives you an equation you can work with, as it has only one nonconstant, i.e. \theta.

In order to find the maximum and minimum values, you must use calculus.

--J
 
Sorry to be so long in responding, I got a nasty flu bug. The above information was just what I needed to break through my mental block, thanks for your prompt and informative response.
 

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