Homework Help: Need help on a max and min force question!

1. Oct 7, 2007

pcandrepair

A block of mass 2.00 kg is pushed up against a wall by a force P that makes a 50.0° angle with the horizontal as shown below. The coefficient of static friction between the block and the wall is 0.258. (The attached picture might help).

A.) Determine the possible values for the magnitude of P that allow the block to remain stationary.
P max =
P min =

Any help would be greatly appreciated. Thanks!

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2. Oct 7, 2007

Staff: Mentor

What forces (vertical and horizontal) act on the block?

3. Oct 7, 2007

pcandrepair

The force in the y-direction(vertical) would be Py = P(sin(50)) and in the x-direction(horizontal), the force would be Px = P(cos(50)). The x-direction would also include the normal force, n, and the y-direction would include gravity*mass, mg.

4. Oct 7, 2007

pcandrepair

I forgot to add that the x-direction would also include the normal force, n, which would be located along the negative x-axis. The y-direction would include gravity*mass, mg, on the negative y-axis.

5. Oct 7, 2007

Staff: Mentor

What about friction? What's its magnitude and which way does it act?

6. Oct 7, 2007

pcandrepair

Friction is the normal*coefficient of friction. I found that the normal force is n = Pcos50. To find the max wouldn't you make friction negative because it would be acting downwards and vice versa for minimum? I just not sure how to put all these into an equation to solve for P. I also know that the acceleration would be zero.

7. Oct 7, 2007

Staff: Mentor

Exactly right.

Just set vertical forces equal to zero and you can solve for P.

8. Oct 7, 2007

pcandrepair

I don't think i'm getting the right answers....So would the equation (for the max) be
( Psin50 - (-friction(Pcos50)) - mg = 0 )? And the Psin50 and mg would cancel out because they would equal zero? I don't think i'm doing this right...

9. Oct 7, 2007

Staff: Mentor

So far, so good--except for that double minus sign. I'd write it like this:
$$P\sin{50} - \mu P\cos{50} - mg = 0$$

Now just solve for P.

Why would they cancel?

10. Oct 7, 2007

pcandrepair

I thought that's what you meant by set the vertical forces to zero because Psin50 and mg are vertical forces. When i solve for Pmax i'm getting 154.282 N. If you made the friction positive to find the min wouldn't that make the force the same but just negative?

11. Oct 7, 2007

Staff: Mentor

But friction is a vertical force also! There are three vertical forces, which add to zero.
I don't see how you got this result.

No.

12. Oct 7, 2007

pcandrepair

If the vertical forces, Psin50, -friction, and -mg add to zero, i don't see how you can solve for P or how it could be a reasonable answer. I think i'm making this harder than it is :S

13. Oct 7, 2007

Staff: Mentor

Solve the equation I gave in post #9. The only unknown is P; everything else is just numbers.

14. Oct 7, 2007

pcandrepair

Psin50 + P(.258)(cos50) - 2kg(9.8) = 0

Psin50 + P(.258)(cos50) = 19.6

2P = 19.6 / (cos50)(sin50)(.258)

P = 154.282 / 2
(I forgot to divide by two last time i got 154.282)

P = 77.141 N

Or am i making some stupid basic math mistake?

15. Oct 7, 2007

Staff: Mentor

That first + sign should be a minus sign.

That should be:
Psin50 - P(.258)(cos50) = 19.6

No good. Factor out the P.

For example, note that Xa - Xb = X(a - b)

16. Oct 7, 2007

pcandrepair

I always seem to make those basic math mistakes! I entered my answers and they were correct. Thank you for all your help Doc Al!

17. Oct 7, 2007

Staff: Mentor

Excellent. Why don't you mark this one solved.

18. Oct 7, 2007

pcandrepair

Actually, the is a part b and it has the same numbers except for an angle of 13 degrees from the horizontal. This time I'm getting a negative number for some reason

19. Oct 7, 2007

Staff: Mentor

If the only difference is that the new angle is 13 degrees (instead of 50), then you can solve it in exactly the same way.

20. Oct 7, 2007

pcandrepair

When i do that i get P = 19.6 / -.02 which turned out to be -714 ish