Coefficient of kinetic friction adding

[A_lilah]

Homework Statement

In the figure below, a box of ant aunts (total mass m1 = 1.15 kg) and a box of ant uncles (total mass m2 = 3.60 kg) slide down an inclined plane while attached by a massless rod parallel to the plane. The angle of incline is θ = 30°. The coefficient of kinetic friction between the aunt box and the incline is µ1 = 0.226. The coefficient of kinetic friction between the uncle box and the incline is µ2 = 0.113.
(a) find the tension in the rod
(b) Compute the common acceleration of the two masses

Homework Equations

Fnet = ma

The Attempt at a Solution

(a) Solved part a and got that the tension was .836722N (which was right)
but I don't think you need that for part b...
(b)
Drew a new FBD for the total system:
W = force down = (mass of m1 + mass of m2)g =46.5975 N
Normal force parallel to the ramp
Frictional force perpendicular to the ramp, going up it

Fnet perpendicular = Normal force - perpendicular component of the weight = N - 46.5975cos(30) = 0 (because it doesn't accelerate in the perpendicular direction)
N = 40.3546

the frictional force = coefficient of friction * normal force

for the coefficient of friction I added the two coefficients given: .113 + .226 = .339 (not sure if this part was right...)

frictional force = 13.6802N up the ramp

so Fnet parallel to the ramp = ma = Parallel weight force - frictional force

Fnet = 46.5975a = 46.5975N * sin(30) - 13.6802

solve for a, which = .7064m/s^2, which wasn't right.

Any help with this second part would be great, thanks!

 

[Dick]

If you got a) right then you clearly know what you are doing. You've highlighted the problem. You can't add the coefficients of friction. Just treat b) as an extension of a) and add your tension force to one of the force diagrams from a).

 

[A_lilah]

oh...

got it :) thanks for your help