A .5 kg block of cheese sits on a table (Static Friction q)

AI Thread Summary
A .5 kg block of cheese on a table experiences static friction with a coefficient of .60, while a mouse hangs from a knot of three strings, one tied to the cheese and another at a 30-degree angle to a wall. The maximum mass of the mouse for equilibrium is calculated to be .16 kg. The tension in the strings needs to be analyzed to understand the forces acting on both the knot and the cheese. The vertical component of the tension must support the mouse, while the horizontal component must counteract the frictional force on the cheese. A free body diagram is essential to clarify the forces and equations involved in this static friction problem.
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Homework Statement



A .5 kg block of cheese sits on a level table. The coefficient of static friction is .60. Three strings are tied together in a knot at a point, K. One string is tied to the cheese, the other on a wall at a 30 degree angle, and in the middle of those two strings a mouse hangs from them. What is the maximum mass of the mouse for the cheese and mouse to remain in equilibrium.

Homework Equations



F = ma Fg = mg

The Attempt at a Solution



So, I calculated the Ff of the cheese, being 2.948 N.
I know I need to calculate the Ft of the left string, and I think that equation would be for Ftx = (mass mouse)(-9.81) / tan30.

I don't know how to calculate Fty, and I don't j=know how to finish the question up. The answer is .16kg.
 
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I assume the string from the cheese to the knot is horizontal. You mention tension in "the string", but there are three strings. Please use distinct symbols for the different tensions. What statics equations do you have at the knot? At the cheese?
By the way, seems to me you are almost there.
 
Since you don't show the diagram, I would assume that the string attached to the cheese is parallel to the table.
If that's the case then the vertical component of the tension in the string (at 30 deg) would have to support the mouse and
the horizontal component of that string would provide the force needed to move the cheese.
It looks like that answer given is slightly less than the mass calculated using this solution.
Perhaps this is because the cheese could move using this value for m.
 
Sorry, the cheese string is in fact parallel. The coefficient of friction between the cheese and the table is .60. I don't get how to find tension in the right string (the 30 degree one) without being given mass, because mass of the mouse determines the tension?
 
Blobikins said:
Sorry, the cheese string is in fact parallel. The coefficient of friction between the cheese and the table is .60. I don't get how to find tension in the right string (the 30 degree one) without being given mass, because mass of the mouse determines the tension?
Draw the free body diagram for the knot. There are three forces on it. What two equations can you write down?
 
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