Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

How to break up a system?

Theres this problem that I have to do:

A friend of mine gave me the answers so im not really interested in the answers but I want to know and understand how to derive those answers? What is approach of a system with multiple objects etc?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: How to break up a system?

A
The whole system is acted on upwards by 200 N and downwards by its weight, which is (6+4+5)g = 15g N. So the resultant upward force on the system is 200−15g N. By Newton’s second law, this is equal to the mass of the system times its acceleration a, i.e.

B
Consider the forces on the 6kg (topmost) block. There’s the 200N force pulling it upwards; pulling it downwards are its weight of 6g N and the tension T at the top of the heavy rope. So the resultant upward force on the 6kg block is 200−(6g+T); by Newton’s second law

(where a is the acceleration found in part (A)).

C
Now consider forces on the 5kg (lower) block and the lower half of the rope. This is being pulled upwards by the tension in the middle of the rope, T′, and downwards by its weight. If we assume the rope is homogeneous, then the mass of half of the rope is 2 kg; hence the weight of half the rope and the lower block combined is 7g N. So the resultant upward force on the lower block and the lower half of the rope is T′−7g, and by Newton

Last edited by JaneFairfax (2008-02-10 09:47:56)