Did I Calculate the Tension in the Cord Correctly?

[apiwowar]

Figure 5-56 shows a box of mass m2 = 1.3 kg on a frictionless plane inclined at angle θ = 34°. It is connected by a cord of negligible mass to a box of mass m1 = 2.7 kg on a horizontal frictionless surface. The pulley is frictionless and massless. (a) If the magnitude of the horizontal force F is 2.5 N, what is the tension in the connecting cord? (b) What is the largest value the magnitude of F may have without the connecting cord becoming slack?


so i broke this up into 2 free body diagrams

for the first one i got T+2.5 = 2.7a so T = 2.7a-2.5

for the second one T-7.1 = 1.3a so T=1.3a+7.1

putting them together and solving for a i got a = 6.86m/s^2

when i plugged that back to solve to T i got T = 16.0N

but its ssaying that 16.0 is wrong, did i do anything wrong?

and can i get a hint for part b?

 

[Ry122]

You forgot to take into account the fact that it's only the vertical component of the weight that's going to affect the tension.

 

[apiwowar]

what do you mean?

 

[Ry122]

Well, imagine if that 34 degree incline was non existent, and the m2 was instead hanging in the air, with no surface supporting it. Don't you think that would change the tension in the cable?

That value of 16.0N that you calculated would be correct if in such a scenario because you haven't taken into account that there is some support being provided to m2.

 

[apiwowar]

if there was no surface then the force downward would just be the weight, but since there is a surface you find the weight that is parallel to that surface which is mgsin(x). is that what youre getting at?

 

[Ry122]

Yeah, whether it's sin or cos I'm not sure, but since you have the final answer already you can figure that one out.