# Determine magnitude and direction of the force exerted on the floor

1. Nov 15, 2012

### cdornz

1. The problem statement, all variables and given/known data
There is a 180 pound basketball player. The coefficient of friction between his right foot and the floor is 0.7. There is no acceleration or sliding in the problem. Determine the magnitude and direction of the force that his right foot is exerting on the floor.

2. Relevant equations
ƩF=0 (know what since acceleration is zero)
Ffriction = (0.7)(Fn) (I know Fn is perpendicular to the interface between the two surfaces and it is not equal to the weight of the object. This is what's stated in my physics book.)
Fright foot = -Fgravity - Fn - Ffriction

I would then use component forms of the equation, so one for "x" and one for "y"

3. The attempt at a solution
My attempt at this isn't coming out correctly and I honestly have no idea why.

(for the x equation) (0.7)Fncos0 = Fright footx
(for the y equation) 800sin270 + Fnsin90 = Fright footy
-800 + Fn = Fright footy

I don't know where to go from here, because I'm pretty sure it's already wrong. Any help would be appreciated!

2. Nov 15, 2012

### haruspex

What makes you think there are any horizontal forces involved?

3. Nov 15, 2012

### cdornz

I had assumed when I was given coefficient of friction, that that was involved then. Also if it isn't, the Fright footx would be zero and that doesn't make sense.

4. Nov 15, 2012

### haruspex

That's a dangerous assumption. Good examiners throw in irrelevant information to test for understanding rather than blindly plugging numbers into equations.
Why not? Would the player be able to stand upright on ice without sliding? What forces would be involved then?

5. Nov 17, 2012

### cdornz

So I had the chance to re-evaluate this problem a little bit and tried solving it from
$\Sigma$F=ma and I know that a=0

so I would have two equations (to get the two components) so one for sine and one for cosine. When I assume that Fn is 180 pounds as well this is what I get:

Fg + Fn + Fright footx = ma
180cos270° + 180cos90° + Fright footx = 0
0 + 0 + Fright footx = 0
Fright footx = 0

Fg + Fn + Fright footy = ma
180sin270° + 180sin90° + Fright footy = 0
-180 + 180 + Fright footy = 0
Fright footy = 0

But, when I don't assume Fn is 180 pounds, this is what I end up with:

Fg + Fn + Fright footx = ma
180cos270° + Fncos90° + Fright foodx = 0
0 + 0 + Fright footx = 0
Fright footx = 0

Fg + Fn + Fright footy = ma
180sin270° + Fnsin90° + Fright footy = 0
-180 +Fn + Fright footy = 0
Fn + Fright footy = 180

In either of these cases, something still doesn't seem right and I'm truly lost as to what component I'm missing.

6. Nov 17, 2012

### ap123

Is there any more to this question that you have left out?
There doesn't seem to be enough information to solve it as it stands.

7. Nov 17, 2012

### cdornz

That was my thought, no there is nothing more to this problem as given. I guess it's possible he left stuff out, but it wouldn't explain how I could solve others in this section.

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8. Nov 17, 2012

### ap123

The picture shows a 180 pound basketball player

The picture might help.

9. Nov 17, 2012

### cdornz

Here's the picture.

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10. Nov 17, 2012

### ap123

OK, here's my analysis of the situation - I'm not sure if it's completely correct, (maybe someone else can check).

The force that his foot exerts on the floor is equal and opposite to the force that the floor exerts on his foot.
The former is the vector sum of the normal force and the frictional force ( which acts to the right in the picture ).
The normal force is equal to his weight (since there is no acceleration in the vertical direction), and the frictional force = 0.7 * his_weight.
Adding these 2 vectors gives a force of magnitude 220lb, acting at an angle of 55° to the horizontal.
So, the force exerted by his foot is equal and opposite to this.

11. Nov 17, 2012

### cdornz

So the answer key shows this is the correct answer, but equation wise could you help me break it down?

So you are saying that the arrow (that would be drawn through his right leg) is the sum of the normal force and frictional force? So if normal force is 180 and frictional force is 126, how would that equal 220? I'm just trying to get as accurate of notes as possible so I know how to approach future questions like this.

12. Nov 17, 2012

### cdornz

Wait, nevermind, my bad. I forgot to apply Pythagorean theorem. Thanks for the help!