Two Masses, a Pulley, and an Inclined Plane

[pkreilley]

Homework Statement

Block 1, of mass m1 = 0.700 kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0 ∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.350, an acceleration of magnitude a = 0.200 m/s2 is observed for block 2.

Find the mass of m2.

Relevant Equations

Fnet=ma
T= mg-ma
[m1]ΣFy=m1g-T
[m2]ΣFy=m2gcosΘ-Fn=0
[m2]ΣFx=m2gsinΘ+μFn-T=m2a

I solved for T on m1 and arrived at 6.72. I plugged that value into the ΣFx equation as shown above (pardon my handwriting) and got a mass of 0.88 kg.

The online program indicated that I needed to check my expression for tension, noting that the two tensions are heading in opposite directions. I don't really understand the significance of that feedback. Any help is much appreciated.

 

[kuruman]

Welcome to PF @pkreilley.

Homework Statement:: Block 1, of mass m1 = 0.700 kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0 ∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.350, an acceleration of magnitude a = 0.200 m/s2 is observed for block 2.

Find the mass of m2.
Relevant Equations:: Fnet=ma
T= mg-ma
[m1]ΣFy=m1g-T
[m2]ΣFy=m2gcosΘ-Fn=0
[m2]ΣFx=m2gsinΘ+μFn-T=m2a

View attachment 269833View attachment 269835
I solved for T on m1 and arrived at 6.72. I plugged that value into the ΣFx equation as shown above (pardon my handwriting) and got a mass of 0.88 kg.

The online program indicated that I needed to check my expression for tension, noting that the two tensions are heading in opposite directions. I don't really understand the significance of that feedback. Any help is much appreciated.

This is where your problem is
[m2]ΣFx=m2gsinΘ+μFn-T=m2a
Up the incline (direction of acceleration a) is assumed positive as shown on the RHS.
m2gsinΘ and μFn must be negative because they are down the incline
T must be positive because it is up the incline.

Or you could just change the sign of the acceleration in this equation and leave the LHS alone.