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

In summary, the mouse weight causes the cheese to move. The mouse is attached to the left string and hangs at a 30 degree angle from the right string. The tension in the left string balances the weight of the mouse.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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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Related to A .5 kg block of cheese sits on a table (Static Friction q)

1. What is static friction?

Static friction is the force that prevents an object from moving when it is in contact with a surface. It is caused by the interlocking of microscopic irregularities on the surface of both the object and the surface it is resting on.

2. How is static friction different from kinetic friction?

Static friction only occurs when an object is at rest, while kinetic friction occurs when an object is in motion. Additionally, the force of static friction is typically greater than the force of kinetic friction.

3. What factors affect the amount of static friction on an object?

The amount of static friction depends on the weight of the object, the roughness of the surface it is resting on, and the force applied to the object.

4. How is the coefficient of static friction calculated?

The coefficient of static friction is calculated by dividing the maximum force of static friction by the normal force (the force perpendicular to the surface) between the object and the surface.

5. How would the static friction on a .5 kg block of cheese change if the surface it is resting on becomes smoother?

If the surface becomes smoother, the static friction on the block of cheese would decrease. This is because there would be less interlocking between the microscopic irregularities on the object and the surface, resulting in a smaller force of static friction.

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