Dynamics (urgently need help!)

Audi SG

Some serious problems with the concepts, and I have difficulties solving even questions most people will consider simple.

Eg 1. A light rope passes over a smooth pulley. 2 weights of masses 4.0kg and 6.0kg are suspended from each end. Determine the

A) acceleration of the 4.0 kg mass and

B) the tension in the rope

Taking upwards positive

A) (6.0 X 9.81) - (4.0 X 9.81) = 10a

a = 1.962 m s- 2 = 2.0 ms-2

Notice that the 9.81 ms-2 used are both +ve. I was told that we treat such equations in forces as summing/subtracting in for scalars. And that if I take downwards as +ve, I have to equate (6.0 X 9.81) - (4.0 X 9.81) = 10 (-a) but I don't understand WHY. Why -a?

B) Let's call the 4kg mass A and the 6kg mass B.

Of course I know tension in both the rope is the same magnitude and in same direction. (please correct if i'm wrong) Which means I only have to solve for 1 Free body diagram to get the tension. But to double check my answer, I did it for the 2 FB diagrams to see if I can values of same magnitude and direction

So for mass A

Ta - Wa = 4 (1.962)

Ta - (4 X 9.81) = 7.848

Ta = 47.1 N (upwards)

When I solve for mass B

I did,

Wb - Tb = 6.0 X -1.962

Tb = (6.0)(+9.81) + 11.772

= 70.something (cant remember the ans)

well I used +ve for 9.81 because I treated this as addition of scalar... but this answer is far from this actual ans of 47.1N and it has LARGELY to do with my hazy concept of determining the direction... can someone explain to me clearly please? best if there's a simple diagram because i'm so troubled that i am having difficulties over supposedly easy questions like this. thanks!

wee_ws SG

Hi,

If the downward direction is taken as positive, we use -a because the 4.0 kg mass is moving upwards.

If we express a in terms of g, then both tension values will be the same.

In (a), we have 2g = 10a => a = 0.2g.

For 4.0 kg mass: T - 4g = 4(0.2g) => T = 4.8g.

For 6.0 kg mass: T - 6g = 6(-0.2g) => T = 4.8g.

Here's a useful link for your reference: http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section2.rhtml

Jiayou!

Cheers,
Wen Shih

Audi SG

Hi,

If the downward direction is taken as positive, we use -a because the 4.0 kg mass is moving upwards.

If we express a in terms of g, then both tension values will be the same.

In (a), we have 2g = 10a => a = 0.2g.

For 4.0 kg mass: T - 4g = 4(0.2g) => T = 4.8g.

For 6.0 kg mass: T - 6g = 6(-0.2g) => T = 4.8g.

Here's a useful link for your reference: http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section2.rhtml

Jiayou!

Cheers,
Wen Shih

Thanks for the help, but I do not understand the part where 2g = 10a.

And I have a question that I have difficulties in obtaining the correct direction, and I kept getting an answer that is actually larger/smaller than the actual value. Here goes:

A life has a mass of 1.2 X 10^3 kg and a man of mass 8okg is standing on a weighing machine of negligible mass inside the life (assume g = 10ms-2) When the lift is movingup with a deceleration of 2.0ms-2, find

the force that the weighing machine exerts on the man.

taking downwards +ve

W(man's weight) - F (weighing machine) = 80 (-2.0)

I got 960 N as the answer. Is the answer supposed to be 0.64 kN? I guess it must be that i interpreted the direction wrongly but I dont see where I have gone wrong.

find also the apparent mass of the man in kg as shwon in the weighing machine

sum of all forces = ma

is the total F = 640 N and a = (10-2.0)? not sure of which value of a to take

Thanks again. :)

wee_ws SG

Hi,

2g = 10a

comes from your original working

(6.0 X 9.81) - (4.0 X 9.81) = 10a.

For the 1st part of your weighing machine problem,

F - 80(10) = 80(-2), since the lift is moving up with a deceleration.

So F = 640 N or 0.64 kN equivalently.

Apparent mass = 640 / 10 = 64 kg.

This document is great at explaining the different scenarios faced by a person on scale in an elevator:

http://physics.usask.ca/~kathryn/phys111/apparent_weight.pdf

Thanks.

Cheers,
Wen Shih