Newton : Tut 3

I am using f to represent friction.

1. A box of weight 15N is placed on a horizontal surface. The coefficients of static friction and kinetic friction are 0,6 and 0,4 respectively.

1.1. Why do these coefficients not have units?
1.1. These coefficients are just ratios, of force divided by force. So units cancel out.
1.2. Show that you need a force greater than 9N to move the box from rest.
1.2. Calculate the magnitude of the static friction fs.






Since the static friction (the force that prevents it from moving) is 9N, you need a force greater than 9N to overcome the friction.
1.3. After the static friction has been overcome, assume that you apply exactly 9N. Calculate the acceleration of the block.
1.3. You need to calculate the kinetic friction now. The effect of the static friction is now over.




You need to calculate the mass of the box since this was not given. Only the weight was given.


1.4. If instead of accelerating the box as in 1.3 assume you wanted to move the box at constant velocity after overcoming the static friction. What force would you need to apply?
1.4. Apply a force of 6N.
Since the kinetic friction is 6N, and the applied force is 6N, the net force would be zero. According to Newton's First Law, the box would move at constant velocity due to it's inertia.
1.5. Assume that you just kept increasing the applied force from zero to 20N.
Draw a graph of Friction Force vs Applied force.
1.5.



2. The driver of a 2000kg car moving at 72 km.h-1, notices a object on the road a distance of 35m in front. The driver presses the brakes. If the coefficient of kinetic friction for the car and the road is 0,6 can the car stop before the object?
2.


Since the vehicle stops in 34,01m, the car does not hit the object which was 35m away.

3. A force of 13N is applied horizontally on a block against a wall. The block has a weight of 5,7N. The coefficient of static friction between the wall and the block is 0,6 and the coefficient of kinetic friction is 0,5.


3.1. Draw a force diagram showing all the forces acting on the block.
3.1.
3.2. Show by means of a calculation why the block is actually at rest.
3.2. Calculate the maximum static friction.


The blocks weight is only 5,7N and is unable to overcome this maximum static friction of 7,8N. Hence the block does not move.
3.3. What is the actual magnitude of static friction keeping the block at rest?
3.3.
5,7N
Static friction can have many values, up to it's maximum. It's maximum in this example is 7,8N. But it is only exerting enough friction force to balance the weight of 5,7N.
3.4. If the applied force is gradually decreased, eventually, the block would slide down.
What is the minimum applied force, that would first allow the block to just start sliding downwards?
3.4. The weight is 5,7N. So the friction must become LESS than 5,7N. Therefore calculate the normal force that would produce a friction force equal to the weight of 5,7N. Then use a force slightly less to get a friction slightly less.



This means that any applied force less than 9,5N would produce a static friction force less than 5,7N. And the block would not be able to be supported, and would slide down.
3.5. Calculate the maximum applied force required to now allow the block to continue sliding downwards at constant velocity.
3.5.
Since the system is now moving, remember to use the coefficient of kinetic friction to calculate friction.
For constant velocity, the vertical forces must be balanced. Hence the friction force must equal to the weight.


Apply a force of 11,4N

If you had applied a force of 11,4N at the beginning, before the block started moving, the system would have remained at rest.
You would use the static coefficient of 0,6 and get a friction of 6,84N which would be larger than the weight. And the block would not be moving!
So you need a little extra force to overcome static friction! Then it is easier to move.





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