MHB What is the Magnitude of Force P to Keep a Block Stationary Against a Wall?

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To keep a block of mass 3.00 kg stationary against a wall, a force P at a 50.0-degree angle must be applied, considering the coefficient of static friction is 0.250. The horizontal forces must balance, leading to the equation -N + F*cos(50) = 0, where N is the normal force. In the vertical direction, the friction force acts downward, and its magnitude is f = μN, which must also be accounted for in the equilibrium equations. Analyzing both the horizontal and vertical forces allows for the determination of the possible values for the magnitude of P. Properly accounting for all forces is essential for solving the problem accurately.
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A block of mass 3.00kg is pushed up against a wall by a force of P that makes a 50.0 degree angle with the horizontal (REFER TO PICTURE). The coefficient of static friction between the block and the wall is 0.250. Determine the possible values for the magnitude of P that allow the block to remain stationary.
 

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You have an applied force which has a component in the horizontal direction. You also have a normal force directed in the opposite direction. (Let's say that N is in the negative x direction and the component of the applied force is in the +x.) The sum of these two forces must be 0 N as there is no motion in this direction. This gives you N in terms of the applied force. Thus [math]\sum F _x = -N + F~cos(50) = 0[/math].

What can you say about the vertical (y) direction? Do the same thing. A lot of stuff will cancel out.

By the way, you missed a force on your diagram. There is a friction force acting downward of magnitude [math]f = \mu N[/math].)

See what you can do with this.

-Dan
 
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