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

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Homework Help Overview

The problem involves a .5 kg block of cheese on a level table with a coefficient of static friction of .60. Three strings are connected at a knot, with one string attached to the cheese, another to a wall at a 30-degree angle, and a mouse hanging from the knot. The objective is to determine the maximum mass of the mouse for the system to remain in equilibrium.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating the force of friction acting on the cheese and the tensions in the strings. There are questions about the setup, including the orientation of the strings and the need for distinct symbols for different tensions. Some participants suggest using statics equations at the knot and the cheese, while others express uncertainty about how to find the tension in the angled string without knowing the mass of the mouse.

Discussion Status

The discussion is ongoing, with participants providing guidance on drawing free body diagrams and writing equations based on the forces acting on the knot. There is recognition that the original poster is close to a solution, but clarity is still needed regarding the relationships between the tensions and the forces involved.

Contextual Notes

Participants note the importance of the coefficient of friction and the angles involved, as well as the implications of the mass of the mouse on the tension in the strings. There is an acknowledgment of the need for a diagram to better visualize the forces at play.

Blobikins
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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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