How Do We Calculate the Weight of Block C in a Frictional Force Problem?

[Redcutter]

Hello, I'm new here. I've browsed before and saw that this is a very helpful community, so I came here for assistance.

Homework Statement

Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 27.5N each, and the coefficient of kinetic friction between each block and the surface is 0.40. Block C descends with constant velocity.

Image attached:
http://session.masteringphysics.com/problemAsset/1038585/38/YF-05-56.jpg

Homework Equations

What is the weight of block C?

The Attempt at a Solution

I've found the first part of this problem, which is the tension holding A and B, which is 11N. Part C has me confused. Here's my attempt:
Block B

∑F_Y=
N_B-m_Bgsin(36.9)=0
N_B=27.5Nsin(36.9) = 16.5115562

f_K= µN_B = .40(16.5115562)= 6.604622479∑F_x=
T₂-T₁-f_K- 27.5cos(36.9)= 0

T₂-T₁-6.604622479- 21.99132811= 0

T₁= 11N

T₂-11-6.604622479- 21.99132811= 0

T₂= 17.59595059N.

Block C

∑Fx= T₂-m_Cg = 0
T₂ = m_Cg = 17.59595059N

Is this correct?

Thanks in advanced.

 

[elegysix]

Since the system is not accelerating, $\sum$F=0.

F1=.4*mg=.4*27.5N

F2=.4*mg*sin(36.9 degrees) = .4*27.5*sin(36.9) N

therefore

F1 + F2 = mg (m = mass of block 3)

.4*27.5(1+sin(36.9)) = mg (the weight of block 3)

 

[jaumzaum]

O M = opposite to the movement

Block a:
F in a = F friction in a = N.u = 27,5u O M

Block b:

F in b = Pb sin36,9° and Pb cos36,9°u O M = 27,5.3/5 and 27,5.4/5.u

F friction = 27,5 . 3,6/5 > 27,5.3/5 so they have same direction

27,5.6,6/5 = C = 5,5.6,6 = 36,3N

 

[Redcutter]

Thanks, guys.

 

[rwisz]

Hello and welcome to the community! It's great to hear that you have found this community helpful.

In regards to your question, your solution for part C appears to be correct. However, I would like to suggest a couple of things to consider in your approach.

Firstly, when finding the weight of block C, you could also use the equation ∑Fy = NC - mCg = 0, since block C is at a constant velocity and therefore the net force in the y-direction is zero. This would give you the same result of NC = mCg = 17.59595059N.

Secondly, in your solution for block B, you have correctly found the tension in the rope holding block A and B to be 11N. However, in your calculation for the friction force, you have used the weight of block B instead of the weight of block A. This would give you a slightly different value for the friction force. It may be a small difference, but it's always good to double check your calculations.

Overall, your solution is correct and shows a good understanding of the concepts of frictional forces and equilibrium. Keep up the good work!