# Linear Dynamics

1. Mar 8, 2015

### Felipe c.

Hello, i was practicing for an exam and i found this problem on my book:

A block of mass 3.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown in the figure. The coefficient of static friction between the block and the wall is 0.244. a) Determine the possible values for the magnitude of P that allow the block to remain stationary b) Describe what happens if P is larger than Pmax. c) Describe what happens if P has a smaller value than Pmin. d) Repeat parts (a) and (b) assuming the force makes an angle of θ = 13.1° with the horizontal, and interpret your results.

I dont understand a) and b)
help is welcome!

2. Mar 8, 2015

### Pierce610

a) Fp = m x g = 3 x 9.8 = 29.4 N
Fp is the force on the block due to gravity and it is also the force that we must balance.
The friction force depends by the normal reaction to the piano that here is zero, because there isn't a vertical component of the force; so I'm almost convinced that the coefficient of static friction must not be considered.
40 ° is the angle between the vertical and P.
The sine of 40° is 0.64.
So P = Fp/sin 40° = 29.4/0.64 = 45.93 N

3. Mar 9, 2015

### Pierce610

Gentle Felipe,
I'm afraid to be wrong.
I have reflected until now on the friction: the friction force isn't a gravitational force; it is correlated to a normal (perpendicular) reaction of a plan to any force applied on it, not necessarely the force due to a weight of a mass on it.
Being a reaction force, it borns at the time a force is applied on a plane, also vertical, and so it strictly depends on it.
Its nature depends on electrical attractive or repulsive force.
So I suggest to modify how I said:
a) Fw = m x g = 3 x 9.8 = 29.4 N
Fw is the force on the block due to gravity and it is also the force that we must balance with the friction force.
The friction force depends by the normal reaction to the piano, P x cos 50° and then the friction force is 0.244 x P x 0.64.
So the equation to solve is Fw=0.244 x P x 0.64, then P =29.4/(0.244 x 0.64) = 188.3 N
That is the force P is about six time the weight force;
if P was vertical, then P should be simply 29.4 N (no effects due to friction); if P was horizontal then P becomes 120,5 N.
Now I hope it is correct and I invite other visitors to signal me if I'm wrong.

4. Mar 10, 2015

### Pierce610

The equation at a) point isn't still right.
Let's make a new reasoning.
We could initially think to start with a simply problem: the static friction coefficient is 1 and the applied force on the block is normal to the plane.
In this condition to this force there is an equal and opposite force of reaction; this last one can produce a static friction that, wanting to equilibrate the weight force, Fw = m x g = 3 x 9.8 = 29.4 N, must be equal and opposite.
The normal reaction should be 29.4 if the coefficient was 1, but being it 0,244, this force is 120.5 N; in fact 120.5 x 0.244 = 29.4.
Now the our force P isn't normal to the plane but it forms angle of 50° with the horizontal.
Then P = normal reaction x cos 50° = 120.5 x 0.64 = 77.12 N
This problem was too difficult to me and I had to think again it many times.
Now I'm sure I've finished with a) point. I dont'realize b) point.

5. Mar 10, 2015

### nasu

There is no figure. Is the angle of 50 below or above horizontal?