Finding magnitude of tension that causes object to slip

[B_Ran]

Homework Statement

A rope attached to a 19.0 kg wood sled pulls the sled up a 19.0° snow-covered hill. A 14.0 kg wood box rides on top of the sled.

EDIT: the actual question to be answered, oops.

What's the magnitude of the tension that will cause the box to slide?

Mass of box (m_b) = 14.0kg
Mass of sled and box (m_s) = 33.0kg
Coefficient of static friction for wood on wood (μ_s) = 0.5
Coefficient of kinetic friction for wood on snow (μ_k) = 0.06
Angle of slope (θ) = 19°

Homework Equations

Fnet = m*a
F_s = μ_sn
F_k = μ_kn

The Attempt at a Solution

[/B]
I've drawn the free body diagrams for both just the box, and for the sled and box combination as they sit on the slope. I've drawn graphs of both and made a chart of the force's x and y components. Getting the relevant forces for each, I've composed two Fnet equations in the x direction.

Equation for net force on the box in the x direction:

F_netbx = m_ba
-Fgsin(19^°) + f_s = m_ba
-m_bg*sin(19) + m_bgμ_s = m_ba
-14(9.8)sin(19) + 14(9.8)(0.5) = 14a
-44.668 + 68.6 = 14a
23.932 = 14a
acceleration = 1.71
I found the acceleration and the answer seems reasonable given the context of the problem.

Equation for net force on the sled+box in the x direction:

F_netsx = m_sa
T + f_s - f_k - F_g1sin(19) - F_g2sin(19) = m_sa
T + m_bgμ_s - m_sgμ_k - m_bgsin(19) - m_sgsin(19) = m_sa
T + 14(9.8)(0.5) - 33(9.8)(0.06) - 14(9.8)sin(19) - 33(9.8)sin(19) = 33(1.71)
T + 68.6 - 19.404 - 44.668 - 105.289 = 56.43
T - 100.761 = 56.43
T = 157.191
Having the acceleration from the first equation, I solved for the tension on the rope. This number concerns me, 157N seems a little low for a force that is pulling that much weight up the slope.

Having used the coefficient for static friction of wood on wood in finding this tension, I was thinking that any force greater than it would cause the box on the sled to slip. My answer of 157.2 is incorrect.

After this I tried going back to just the box.
If f_smaximum = μ_s*n
f_smaxmimum = (0.5)(14)(-Fgcos(19))
f_smaxmimum = (0.5)(14)(-14*9.8*cos(19))
f_smaxmimum = 908
This number doesn't make any sense to me and its clear now that I'm lost. :]

Did I make a mistake somewhere along the lines, or have I not taken this problem far enough? Thank you in advance.

 

[kuruman]

The normal force on the box is not the weight of the box. That would be the case on a horizontal surface, not on an incline.

On edit: Welcome to PF.

 

[B_Ran]

The normal force on the box is not the weight of the box. That would be the case on a horizontal surface, not on an incline.

On edit: Welcome to PF.

Thank you for the welcome!
I considered this and thought I had done right when instead of multiplying the mass of the object by just gravity, as is the case when the object is on a flat surface, I multiplied the mass by the y component of its weight. I see now where I went wrong because that value still isn't parallel with the normal force. I'll attempt again with this information thank you

 

[kuruman]

Equation for net force on the sled+box in the x direction:
T + f_s - f_k - F_g1sin(19) - F_g2sin(19) = m_sa

There is more. Lose the f_s in this equation. If you are writing the net force on the two-mass system the force of static friction on the wood is equal and opposite to that on the sled so they add up to zero.