What is the Maximum Angle for a Crutch to Not Slip on the Ground?

[----md]

Hi All- Hope you can help me :|

Homework Statement

The person in the drawing is standing on crutches. Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing indicate. If the coefficient of static friction between a crutch and the ground is 0.90, determine the largest angle θMAX that the crutch can have just before it begins to slip on the floor.

Homework Equations

Fup= Fdown
Fright= Fleft
Fnet=0

The Attempt at a Solution

I figured the for each crutch
F_y = N= 1/2mg + FcosΘ
F_x= μ_sN=FsinΘ
But I end up with mg and no way to eliminate it?

I know I am missing some important detail- but as with many force problems, I can't visualize it.

Thanks

 

[goraemon]

It appears from the picture that the person is standing on one of his legs in addition to the two crutches. So wouldn't there be 3 upward forces: normal force on his leg, and 2F cos $\theta$?

 

[----md]

Does that mean that because he is standing- his weight isn't a factor in the upwards forces- because it gets canceled out? and the only net force is the vertical component of the crutch?

 

[goraemon]

Upon thinking some more, assuming that he's supporting himself on one leg plus 2 crutches doesn't seem to help. For the moment, let's assume that he's supporting himself entirely on the two crutches only. With this assumption, let's take a look at the vertical net force:

Fnet(y) = 0 = 2F cos $\theta$ - mg

Now, I'm unsure how you derived the following: Fy = N= 1/2mg + FcosΘ. Can you explain your thought process?

 

[----md]

I realized my mistake. SO I was assuming that the F of the crutches are like an external force. But its just a component of the mg. So

I set it up as μmgsinθ=mgcosθ -> .9 tan^-1= θ

Thanks.

 

[goraemon]

I realized my mistake. SO I was assuming that the F of the crutches are like an external force. But its just a component of the mg. So

I set it up as μmgsinθ=mgcosθ -> .9 tan^-1= θ

Thanks.

Just a sec, $.9 tan^{-1}= θ$ isn't a valid equation. Do you mean $tan^{-1}(.9)=\theta$?