Help with a Friction Question: Maximum mass a Man can Push

[jessicak]

Homework Statement

A man of 72.0kg is pushing a heavy box along a flat floor. The coefficient of sliding friction between the floor and the box is 0.20, and the coefficient of static friction between the man's shoes and the floor is 0.65N.

a) If the man pushes downward on the box at an angle of 30 degrees, what is the maximum mass of the box be can move?

b) If the man pushes upward on the box at an angle of 30 degrees, what is the maximum mass of the box he can move?

Homework Equations

f_k= u_kN

The Attempt at a Solution

So, according to the free body diagram for the man:
F_{x man}= -u^km_mang-Pcos(60)
For the box:
F_{x box}= Pcos(30)-u_km_boxg

I don't know where to go from here. If I set the equations equal to one another I have two unknowns (P and m_box). If I add the equations, I get the same problem, plus acceleration thrown in there. If anyone can steer me in the right direction, that would be great!

 

[fzero]

Forget about the box for a second and imagine just a force acting on the man. How large does the force have to be before he begins to slide along the floor?

 

[jessicak]

The force should be equal to the force due to friction of the man?

 

[fzero]

The force should be equal to the force due to friction of the man?

Right, it's the force due to static friction on the man. Can you see how this relates to the original question about pushing the box?

 

[jessicak]

Right, it's the force due to static friction on the man. Can you see how this relates to the original question about pushing the box?

Not exactly. If the force is equal to the static friction of the man, where does the angle of the force come into play? I've spent 3 days on this problem and I am a little lost and frustrated.

 

[fzero]

Not exactly. If the force is equal to the static friction of the man, where does the angle of the force come into play? I've spent 3 days on this problem and I am a little lost and frustrated.

As for the angle, in the equations you wrote down, you neglected to work out the vertical forces on the box. The normal forces involve more than just the force of gravity. Note that this also affects the force due to static friction on the man.

 

[jessicak]

Ok, so the vertical equations are:

F_{y man}= N-m_mang + Psin(60) =0
F_{y box}= N-m_box - Psin(30) =0

If I sub in the values of N, I'm still left with two variables, P and m_box

 

[fzero]

Ok, so the vertical equations are:

F_{y man}= N-m_mang + Psin(60) =0
F_{y box}= N-m_box - Psin(30) =0

If I sub in the values of N, I'm still left with two variables, P and m_box

You should really write N_man and N_box. Now we consider the horizontal force on the man. There is static friction, so as long as P cos(60) does not exceed u_sN_man, the man does not slide. The value P_max at which he does begin to slide is the maximum value of the force he can exert on the box. The horizontal force equation on the box will let you solve for m_max.

 

[jessicak]

Ok-

So F_{x man}= -u_s(m_mang - Psin(60)) + Pcos(60)= 0 (man not moving) then P=u_sm_mang/(cos(60)+u_ssin(60) and P=432N

Can I just use this P value in the F_{x box} and assume the equation is zero?

 

[fzero]

Ok-

So F_{x man}= -u_s(m_mang - Psin(60)) + Pcos(60)= 0 (man not moving) then P=u_sm_mang/(cos(60)+u_ssin(60) and P=432N

Can I just use this P value in the F_{x box} and assume the equation is zero?

Yes, you can set F_{x box}=0 because we're looking for the point where the acceleration becomes non-zero. Just make sure you use the proper N_box depending on whether the man is pushing up or down.

 

[jessicak]

Using P=-432N gives me an m_box of 213kg, which my homework says is wrong

I'm using m_box= (u_kPsin(30) + Pcos(30))/(-u_kg)

 

[fzero]

Sorry, I hadn't been following the algebra closely, I was mainly trying to give hints on the concepts. When the man is pushing down, I find

F_{\text{man},y} = N_{\text{man}} + P \sin(30^\circ) - m_{\text{man}} g

F_{\text{man},x} = - P \cos(30^\circ) + u_s N_{\text{man}}F_{\text{box},y} = N_{\text{box}} - P \sin(30^\circ) - m_{\text{box}} g

F_{\text{box},x} = P \cos(30^\circ) - u_k N_{\text{box}}

Then I find

P_{\text{max}} = \frac{m_{\text{man}} g}{\frac{\cos(30^\circ)}{u_s} + \sin(30^\circ)}

and

m_{\text{box, max}} = \frac{\frac{\cos(30^\circ)}{u_k} - \sin(30^\circ)}{\frac{\cos(30^\circ)}{u_s} + \sin(30^\circ)} m_{\text{man}}.

I can't guarantee that there are no sign errors or typos anywhere, but hopefully it helps.

 

[jessicak]

Yes! I got it. It ended up being a small geometry error in my free body diagram that was my downfall. THANK YOU for all your help, I completely understand where I went wrong and know how to fix it!