Hi socialdiscord!!
I will start with an important definition, the Normal Force:
When an object is resting on a horizontal surface it feels the force
of gravity pulling toward the Earth. But the object is not
accelerating so there must be an opposing force acting on it, this
force is caused by the supporting surface and is known as the Normal
force. This force arises from the repulsive forces between the atoms
at the surface of the object and the atoms at the supporting surface.
The normal force is perpendicular to the surface that causes it, and
for an object sitting on a horizontal surface, without other forces
involved, the normal force will be equal to the weight of the object.
So the force that must be found in the exercises is the normal force.
See for more references:
"Normal Force":
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/Forces/Normal.html
a) A 15 lb. block rests on the floor. What force does the floor exert
on tbe block?
In this case the normal force (Nf) is equal to the weight (W) of the block.
I will assume that the 15 lb is the weight of the block.
Nf = W = 15 lb.
If 15 lb is the mass of the block:
15 lb is equivalent to 15 * 0.4536 kg = 6.804 kg
W = m * g = 6.804 kg * 9.81 m/s^2 = 66.747 N
Then
Nf = W = 66.747 N
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b) If a rope is tied to the block and runs vertically over a pully,
and the other end is attached to a free hanging 10 lb weight, what is
the force of the floor on the 15 lb block?
In this case you are pulling the block vertically with a force (T1) of
10 lbf trying to rise it. The applied force is not enough to elevate
the block because the weight of the added block is lower than the
weight of the original block. Again the original block is not
accelerating, then the vertical forces are in equilibrium:
W = Nf + T1
Then the normal force (or the force exerted by the floor) is:
Nf = W - T1 = 15 lb - 10 lb = 5 lb
Again if 15 lb is the mass of the block:
W = 66.747 N
1 lbf = 0.4536 kgf
then
10 lbf = 4.536 kgf
1 kgf = 9.81 N
then
10 lbf = 4.536 kgf = 4.536 * 9.81 N = 44.498 N
then
T1 = 10 lbf = 44.5 N (again less than W, so we can apply the same solution)
Nf = W - T1 = 66.747 N - 44.498 N = 22.249 N
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c) If we replace the 10lb weight with a 20lb weight, - now what is the
force of the floor on the block?
In this case the new vertical force (T2) is greater than the weight of
the block, then when this new force is applied the block rises from
the floor and the force of the floor on the block dissapears (Nf = 0
N) .
Note that in this case the resultant vertical force (Fv) is:
Fv = T2 - W = 20lb - 15 lb = 5lb
This Fv accelerate the block verticaly with an acceleration A:
A = Fv / m , where m is the mass of the block.
If 15 lb is the mass of the block:
W = 66.747 N
T2 = 2 * 10 lbf = 2 * 44.498 N = 88.996 N (greater than W, so the same
solution is found, Nf is null).
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I hope this helps you.
If you need a clarification please use the request of answer
clarification feature before rate this answer.
Best regards.
livioflores-ga |
