Solving Moving Crate Problem: Find Pmin

[MorehouseM11]

Homework Statement

A crate of weight w initially lies at rest on a horizontal floor. You then push on the crate with a constant force of magnitude P that is directed at angle $$\theta$$ downward from the horizontal. The coefficient of static friction between the crate and floor is $$\mu$$_s.

(a) Show that, in order to get the crate moving, you must have P $$\geq$$ P_min, where

P_min = $$\mu$$_sw sec$$\theta$$
1- $$\mu$$ _s tan $$\theta$$

Hint: Determine the hardest you can push for the crate to remain at rest

Homework Equations

$$\Sigma$$f_x = ma_x
$$\Sigma$$f_x = ma_y

The Attempt at a Solution

After my free body diagram I have the follwoing:
N - Positive Y Direction
$$\mu$$_s - Negative X Direction
mg - Negative Y Direction
P - Negative X with cos$$\theta$$



$$\Sigma$$f_x = ma_x
-M_s-Pcos$$\theta$$ = ma (i)

Ef_y = ma_y
Psin$$\theta$$ + N - mg = 0 (ii)

Solving for N we get N = -Psin$$\theta$$ + mg
Plug N into Equation (i)
-$$\mu$$_s (-Psin$$\theta$$ + mg) - Pcos$$\theta$$ = ma

I know I am supposed to solve for P but with two P's in my final equation I'm not sure how>
I just need some guidance on what's the next step of this problem. Any help is appreciated.


P.s. Excuse The E it's equal to sigma and the same goes for M it's equal to Muse of I was having trouble getting those symbols to work

 

[anathema]

.

To solve for P, you can substitute the expression for N into the equation for \Sigma F_x, giving you:

-\mu_s(-P\sin\theta + mg) - P\cos\theta = ma

Next, you can factor out the P and solve for it:

P(-\mu_s\sin\theta - \cos\theta) = ma + \mu_smg

P = \frac{ma + \mu_smg}{-\mu_s\sin\theta - \cos\theta}

Now, to determine the minimum value of P, you need to consider the case where the crate is just on the verge of moving, meaning that the acceleration is zero. This means that ma = 0. Substituting this into the equation for P, you get:

P_{min} = \frac{\mu_smg}{-\mu_s\sin\theta - \cos\theta}

You can simplify this expression further using trigonometric identities, specifically:

-\mu_s\sin\theta - \cos\theta = -\mu_s\frac{\sin\theta}{\cos\theta} - \cos\theta = -\mu_s\tan\theta - \cos\theta

Substituting this into the expression for P_{min}, you get:

P_{min} = \frac{\mu_smg}{-\mu_s\tan\theta - \cos\theta}

Finally, you can use the identity \sec\theta = \frac{1}{\cos\theta} to get the final expression for P_{min}:

P_{min} = \frac{\mu_smg}{\sec\theta - \mu_s\tan\theta} = \frac{\mu_smg}{\frac{1}{\cos\theta} - \frac{\mu_s\sin\theta}{\cos\theta}} = \frac{\mu_smg}{\frac{1 - \mu_s\tan\theta}{\cos\theta}} = \frac{\mu_smg\cos\theta}{1 - \mu_s\tan\theta}

Thus, the minimum force required to move the crate is given by:

P_{min} = \frac{\mu_smg\cos\theta}{1 - \mu_s\tan\theta}