Frictional forces between ropes

[tcd89]

Blocks A, B, and C are placed as in the figure (Intro 1 figure) and connected by ropes of negligible mass. Both A and B weigh 23.2 N each, and the coefficient of kinetic friction between each block and the surface is 0.31. Block C descends with constant velocity.
Find the tension in the rope connecting blocks A and B.
What is the weight of block C?
If the rope connecting A and B were cut, what would be the acceleration of C?

$$\sum$$F_x=T₁sin36.9-$$\mu$$
$$\sum$$F_y=T₁cos36.9-w

I tried to find the tension between A and B, but it turned out wrong. I think there is something wrong with the equation I used. =/
And I'm completely at a lost as to how you answer the other two questions.
Help is appreciated! Thanks! =)

 

[LowlyPion]

Welcome to PF.

For the first question the key is that it is moving at constant velocity. Hence Ma is only being pulled with Tension that equals the frictional force retarding it which will be μ*Ma*g isn't it?

If you cut the rope, then that force would then be available to accelerate both blocks wouldn't it?

 

[nlightner]

I would first start by analyzing the given information and drawing a free body diagram to better visualize the forces acting on the blocks. From the diagram, we can see that there are three main forces acting on block C: its weight (mg), the tension force in the rope connecting A and C (T1), and the frictional force (Ff) between block C and the surface.

To find the tension in the rope connecting A and B, we can use the equation \sumFx=0, since there is no acceleration in the horizontal direction. This gives us T1sin36.9-N\mu=0, where N is the normal force between block C and the surface. We can solve for T1 and substitute in the given values to find the tension in the rope connecting A and B.

To find the weight of block C, we can use the equation \sumFy=0, since there is no acceleration in the vertical direction. This gives us T1cos36.9-mg=0, where mg is the weight of block C. Again, we can solve for the weight of block C by substituting in the given values.

If the rope connecting A and B were cut, the only force acting on block C would be its weight (mg). This means that the acceleration of block C would be equal to the acceleration due to gravity (9.8 m/s^2) in the downward direction.

In terms of your attempted solution, it seems like you may have mixed up the values for the frictional force and the normal force. The equation for the frictional force is Ff=\mu N, where \mu is the coefficient of kinetic friction and N is the normal force. Also, when solving for T1, make sure to include the weight of block A since it is also connected to the rope. I hope this helps!