Help with tension problem? I can't find information on this!?

Question

I can't seem to find any information on this in the book, nor do I feel like I really understand the concept. Any help would be greatly appreciated!!!

1. Block m1 has a mass of 4 kg and m2 has a mass of 2 kg. The coefficient of kinetic friction between m2 and the horizontal plane is 0.5. The incline plane is frictionless. The angle of the incline plane is 30 degrees.
a) Calculate the tension in the cord.
b) Calculate the acceleration of the blocks.

Answer 1

It's hard to decode what's going on here. It would really help if there were some diagram, or more detailed description.

I'm GUESSING that we have two blocks connected by a cord, and the cord runs over a frictionless pully, and one block lies on a horizontal plane and the other on an inclined plane. So the idea is that, as the one block slides downhill, it will pull the other block too.

In problems of this type, always look at the total forces acting on each piece. First, look at m1 (ignore m2), and make a list of the forces acting on it (use numbers & variables where appropriate). Then, look at m2 (ignore m1) and make a list of the forces acting on IT.

Two things will help you join these quantities together:
1) The amount of tension T pulling on m1, is the same as the tension pulling on m2 (this is because the tension in a cord is the same everywhere)
2) The acceleration of m1 equals the acceleration of m2 (because they're connected).

So:

Forces on m1 (Block on inclined plane):
----------------------------------------------------------
Weight = m1(g) [down]
Tension = T [diagonally up]
Normal force F1_normal [diagonally up, perpendicular to T]

It's convenient to break the weight into components which are parallel and perpendicular to the incline. In that case, we have:

Total "parallel" forces:
T [upslope]
m1(g)(sin30) [downslope]

Total "perpendicular" forces:
F1_normal
m1(g)(cos30)

The perpendicular forces cancel each other out (we know this because there is no acceleration in that direction). So really the magnitude of the NET force on Block 1 is:

F1_net = m1(g)(sin30) - T

Forces on m2 (Block on horizontal plane)
----------------------------------------------------------
Weight = m2(g) [down]
F1_normal [up]
Tension = T [to the left(?)]
Friction = (0.5)F1_normal [to the right(?)]

For m2, the up & down forces cancel out (because no acceleration in that direction). This means F1_normal = m2(g); and this means Friction = (0.5)m2(g).

So the magnitude of the net force on m2 is:

F2_net = T - (0.5)m2(g)

Now, here's where we use the fact that the acceleration of m1 is the same as the acceleration of m2:

a = F1_net / m1 = (m1(g)(sin30) - T) / m1 = g(sin30) - T/m1
And:
a = F2_net / m2 = (T - (0.5)m2(g)) / m2 = T/m2 - 0.5(g)

Since those two are equal, we have:

g(sin30) - T/m1 = T/m2 - 0.5(g)

Now just plug in the given values for m1 and m2 and solve for T. That answers part (a).

Once you know T, you can plug it into either of the equations for acceleration above. That answers part (b).

Answer 2

your question dent give any clue of a cord ....and from where it is coming from. Your question is incomplete and so cant be answered. you must follow "CONCEPT OF PHYSICS", VOL-1, By H.C. Verma

Answer 3

complex task. query on to google or bing. it can assist!

Answer 4

seeing as your problem doesn't mention a cord....i have no idea