Question

There is a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by man himself. If the man manages to keep the box at rest, what is the weight shown by the machine? What force should he exert on the rope to get his correct weight on the machine?

Answer 1

OMG! Those people creating these questions have such an imagination!

Answer 2

W = ma where a is acceleration due to gravity 9.8 m/s^2
for the weighing machine to correctly measure the weight, the man must put the same amount of force exerted by his weight and the weight of the box
W = 9.8(60. + 30.)

Answer 3

the concept here is for the elevator to remain at rest, so man's weight does not change as if he were standing on a ground.

Answer 4

in essence, for the elevator to hold the man's weight, the elevator must support 588N force for it to remain at rest.

Answer 5

588 would be man's weight

Answer 6

Ok, let's take g = 10 N/kg to make things simple.
@Akhilender
Q1: what is the 1st thing to do when you see such a mechanics problem?

Answer 7

decide the system and make free body diagram

Answer 8

Well done! Answer 1 was:
A1: you need to think about the different systems (if more than one) to which you can apply Newton's laws.
Q2: what system(s) can you use here?

Answer 9

any one of the box , the weighing machine or the man ,may be taken as system

Answer 10

True, but some of them are not really useful. And you did not mention their possible combinations.
A hint is that the mass of the scales is not given, only the overall mass of the box including the scales.

A2: useful systems are system 1: man alone, or system 2: {man + box + scales}
Q3: How would you write equilibrium of system 2?

Answer 11

T = N + (60+30)g
where T is the tension in the rope and N is the normal force exerted by the wt. machine on the box.

Answer 12

Considering system 2, N is an internal force, so need not be taken into account.
Notice also that the rope is attached twice to the system.

That should help you write the correct expression.

Answer 13

2T = (60+30)g
is it correct now??

Answer 14

Great!
So T = (M+m) g/2 = 450 N

Q4: Now, how do you write equilibrium of man inside (system 1)?

Answer 15

N + T = 60g
N is the normal force exerted by the box on the man.

Answer 16

Perfect. Now solve for N and the fist question is over.

Answer 17

N = 150
mg=150
m=15kg

Answer 18

Ok, now, second question.
Do you agree that the system will accelerate, and that all bodies will accelerate at the same rate, that we can call \(a\) ?

We will be faces with 3 equations:
- N's 2nd law for system 2
- N's 2nd law for system 1
- imposed condition that scales' reading is man's weight.

Can you write those 3 equations (Please keep letters m, M, T, g, a). Of course T will have a different value from that it had in the first question.

Answer 19

*faced

Answer 20

If we want to write N's 2nd law for system 2, we need to list the forces acting and to know the mass of the system. That's all.
Can you do this?

Answer 21

ok
1) 2T = (M+m)(a+g)

Answer 22

Correct :)
I wrote: 2T - (M+m)g = (M+m)a which is exactly the same.
Go on!

Answer 23

ok 2nd is:
T+N=mg+ma

Answer 24

Ok if small m is the man's mass.
Final equation is the easiest one.

Answer 25

N=mg
here m is the man's mass

Answer 26

Perfect! You did it!
Now you can solve for whichever unknown you want.

Answer 27

T = 1800 N
Which makes a = 30 m/s², which is extremely intense and cannot be sustained for more than a few fractions of second.

Answer 28

Moral of the story:
- better think the problem round before writing down equations
- it is easier to go with small steps that you are sure about, than by a giant leap that leaves you unaware of hidden mistakes.
I hope this will help you deal better with other problems.

Answer 29

thanx for the help @Vincent-Lyon.Fr
I'll keep these things in my head

Answer 30

You're welcome.