I don't even know where to start! :S Here's the question:
Two identical boxes are connected by a string of negligible mass, which runs across a frictionless peg. One box is on a horizontal table, while the other box hangs vertically. If the hanging box is released from rest, it remains at rest.
What must be the force of static friction between the box on the table, and the surface of the table? Assume the boxes each have masses of 2.0 kg, the coefficient of static friction is µs between table and box, and the acceleration due to gravity is 10 m/s^2.
1. Impossible to say from the information given.
2. The force of the static friction would have a vertical component in this situation, and that is impossible since it must be parallel to the table surface.
3. 10 N
4. 5.0 N
5. fs would have to be greater than 20 N, which is physically impossible, so this is not a realistic situation.
6. 20 N
7. 2.5 N
8. Very small, almost zero
Two identical boxes are connected by a string of negligible mass, which runs across a frictionless peg. One box is on a horizontal table, while the other box hangs vertically. If the hanging box is released from rest, it remains at rest.
What must be the force of static friction between the box on the table, and the surface of the table? Assume the boxes each have masses of 2.0 kg, the coefficient of static friction is µs between table and box, and the acceleration due to gravity is 10 m/s^2.
1. Impossible to say from the information given.
2. The force of the static friction would have a vertical component in this situation, and that is impossible since it must be parallel to the table surface.
3. 10 N
4. 5.0 N
5. fs would have to be greater than 20 N, which is physically impossible, so this is not a realistic situation.
6. 20 N
7. 2.5 N
8. Very small, almost zero
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if the hanging mass does not accelerate, then the tension in the string must equal its weight, so we know the tension in the string is 2kg x 10m/s/s = 20N
if the string is massless, the tension in the string is constant throughout, and therefore this is the tension acting on the box on the table
if the box on the table is not accelerating, the friction force must equal the tension tending to pull it, so we can deduce the force of friction is also 20N and option 6 is correct
if the string is massless, the tension in the string is constant throughout, and therefore this is the tension acting on the box on the table
if the box on the table is not accelerating, the friction force must equal the tension tending to pull it, so we can deduce the force of friction is also 20N and option 6 is correct