Finding Mass of Block from Friction Coefficient

AI Thread Summary
The problem involves calculating the minimum mass of a block on a 20-degree slope, connected to a 2.0 kg hanging block, using friction coefficients of 0.80 (static) and 0.50 (kinetic). The forces acting on the block include gravitational force, normal force, and static friction. The correct equation for equilibrium is mg sin(20) + mg cos(20) * 0.8 = 19.6N. After correcting the calculations, the minimum mass m is determined to be approximately 1.83 kg. The solution highlights the importance of accurately applying the forces involved in the system.
merzperson
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*SOLVED*

Homework Statement



Figure shows a block of mass m resting on a 20 degree slope. The block has coefficients of friction \mus=0.80 and \muk=0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

knight_Figure_08_36.jpg


What is the minimum mass m that will stick and not slip?

Homework Equations



Fs = \mus * mg
F = ma

The Attempt at a Solution



The force of the 2kg block on the rope is 2 * 9.8 = 19.6N

Since the m block is not moving, this means that the 2kg block is also not moving, so the force of the rope on the 2kg block is also 19.6N to make the net force 0.

This also means that the force of the rope on m block is 19.6N

I know that there are 3 unknown forces on m block: Earth on the block, normal force, and static friction. The y-component and normal force cancel out so we are left with the x-component of the force of the Earth on the block and the friction force.

mg sin(20) - mg cos(20) * 0.8 = 19.6N

Solve for m. I got m=4.88kg which was wrong. I am lost as to where I might have messed up. It seems like I'm making this problem more complicated than it is.
 
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merzperson said:
mg sin(20) + mg cos(20) * 0.8 = 19.6N


m=1.8267??
 
You got it my man.
 
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