Conceptual gravity question

  • Thread starter tnutty
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  • #1
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When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point "break loose" and start to accelerate down the board.

The box begins to slide once the component of gravity acting parallel to the board equals the force of static friction. Which of the following is the most general explanation for why the box accelerates down the board?

1 The force of kinetic friction is smaller than that of static friction, but remains the same.

2 Once the box is moving, is smaller than the force of static friction but larger than the force of kinetic friction.

3 Once the box is moving, is larger than the force of static friction.

4 When the box is stationary, equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction.


I am thinking its 1.
 

Answers and Replies

  • #2
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Which of the following is the most general explanation for why the box accelerates down the board?
None explains acceleration.

It accelerates because gravity accelerates all objects in proportion to their mass....
 
  • #3
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none is not a option
 
  • #4
Mapes
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Options 2, 3, and 4 aren't sentences and aren't understandable. Is a word missing?
 
  • #5
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Options 2, 3, and 4 aren't sentences and aren't understandable. Is a word missing?
Is this a ploy to pass native speakers?
 
  • #6
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None of those made any sense to me.

This is why the box suddenly accelerates down the board:

The force of gravity is always acting on the box. Since the box is on a tilted board, one component of the gravitational force vector is pulling the board down the slope, and the other is into the board, determining the normal force. When the box is "sticking", it is because the static coefficient of friction between the two objects multiplied by the normal force is greater than the force pulling the box down the slope. As you tilt the board, the component of the gravitational force vector that goes toward pulling the box down the slope becomes bigger. At the same time, the component that determines the normal force becomes smaller. Eventually, the static coefficient of friction times the normal force will equal, and then be smaller than, the force pulling the box down the slope. The static friction will "break" and kinetic friction takes over. However, the kinetic coefficient of friction is always smaller than the static coefficient of friction, so the frictional force is not nearly enough to counter the force pulling it down the slope, so the net force is down the slope and the box accelerates.
 
  • #7
The reason it doesn't make any sense is probably because there is a term missing in every one. It should be:

The actual text is:

1. The force of kinetic friction is smaller than that of static friction, but F_g remains the same.
2. Once the box is moving, F_g is smaller than the force of static friction but larger than the force of kinetic friction.
3. Once the box is moving, F_g is larger than the force of static friction.
4. When the box is stationary, F_g equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction.

---
At the point when the box finally does "break loose," you know that the component of the box's weight that is parallel to the board is equal to mu_s*n (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of mu_s*n , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board...
 
  • #8
327
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... and once static friction is broken, kinetic friction takes over, for which the coefficient is always smaller, so their is a net force, down the board, causing the box to accelerate.
 

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