Calculating Friction and Tension in a Pulley System

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SUMMARY

The discussion focuses on calculating friction and tension in a pulley system involving two boxes with masses of 100kg and 55kg, respectively. A force of 200N is applied at an angle of 50 degrees to the horizontal, moving the boxes 305cm in 5 seconds. The participants concluded that the frictional force can be determined by equating the forces acting on both boxes, while the tension in the rope requires additional information about the system's acceleration, which was calculated to be 0.24m/s² based on the assumption of starting from rest.

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Homework Statement



1. a box has a mass of 100kg, and the second has a mass of 55kg. there is a rope between them that attaches them. there's a rope connecting to the first box, that is pulled with a force of 200N at 50 degrees above the horizontal. the box os moved 305cm for 5s
a) what is the frictional force on the boxes
b) what is the tension in the rope joining the two boxes

2.
a rope is hanging off the side of the table. the rope is 2m long. the coefficient of friction between the table and the rope is 0.62
what is the maximum length of the rope that can hang over without slipping?

Homework Equations



1. i tried using d=vy1*t+0.5g*t^2 to find a. if i assumed vy1 is 0 because it starts from rest. then tried using Ef=Fappy-Ffs to find the mui in FFs...

2. Have no idea how to start

The Attempt at a Solution



have no idea o-o'' what to do ><"
 
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As you've written it, the first problem can't be solved because there isn't enough information. Please tell us the problem statement is exactly as it was given to you.

With force problems, the first thing you should do is identify all the forces acting and draw a free-body diagram for each object.
 
that's actually all the information given :s
that's why I'm findin it hard to do. but i found out how to do a).
to find friction:
equate the two equations
mass A: Ef=Fapp-FT --> Ft= Fapp-Ef
mass B: Ef= Ft-Ffs --> Ft= Ffs + Ef
therefore:
Ffs+Ef = Fapp-Ef
Ffs= Fapp-Ef-Ef(second one)

acceleration found with d=v1y*t + 0.5g*t^2 = 0.24m/s^2
...but idk how to do b) :s

& I'm finding it hard to imagine the rope as a free body diagram.. cause half of its on the table and the other half is off.
so like half of it is lieing horizontally and the other half it lieing vertically.. i can't imagine what force is acting on what part... :s
 
I think in relaying the question here, you've changed the wording and ended up inadvertently omitting some necessary info. That's why I asked you to tell us what the problem statement is exactly as written. Your solution doesn't make sense to me, and as you've written the problem, it can't be solved without making arbitrary assumptions.
 
kkk...

a person pulls a string of two boxes across level ground. the first box has a mass of 100kg and the second, 55kg. pulling the rope attached to the first box with a force of 200N at 50 degrees above the horizontal. the box os moved 305cm for 5s
a) what is the frictional force acting on the boxes
b) what is the tension in the rope joining the two boxes
 
thats it^ the entire question.
 
You can't solve that problem. You don't know the initial velocity, if the boxes accelerate, etc.
 
that's why we assumed that the inital velocity starts from rest, and found the acceleration to be 0.24m/s
it might not be right, but that's what most of us had done.
 

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