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### Course: Physics archive>Unit 3

Lesson 7: Treating systems

# Treating systems (the easy way)

David shows the easier way to find the acceleration of two masses connected by a rope. Created by David SantoPietro.

## Want to join the conversation?

• Why do we consider only the forces acting on the system which are external because even the internal forces of the system affect the motion?
• the internal forces can effect rotation but not the linear motion.
if you move around inside a space ship, it will continue in the same direction, but you can make it rotate
• Shouldn't the Fg be a negative force as well because it pulls downward?
• now in this method the convention changes the forces responsible for accelerating the system are taken as positive
• whats the tension of the rope
• The next video solves for tension, so definitely don't miss that. For this one, we just use Newton's second law for the 5 kg or the 3kg system, your choice. In other words:

(for 5 kg system)
F = ma
T = (5)(3g/8)
=> T = 15g/8

(for 3 kg system)
T - mg = -ma (don't forget minus sign!)
T = -ma + mg
T = m(g - a)
T = (3)(g - 3g/8)
= (3)(5g/8)
= 15g/8
=>T = 15g/8

As you can see, you can do it both ways. Cheers.
• In the problem described in this video and the previous, why doesn't the pulley contribute to the tension of the rope? Wouldn't the normal force of the rope on the pulley effect S(igma)Forces?
• In real life, the pulley would exert some force (e.g. friction). However, we are told in the beginning that this is a weightless and frictionless pulley - therefore we ignore it. We also ignore other forces such as air resistance.
• At the end of the video he finds the acceleration to be 1.86 or something and in the previous video, it was 3.68 why is that although the two diagrams are exactly the same?
• That's because in this video, he added the force of friction to the system, whereas in the previous video, he assumed that friction was negligible. Hope that helps
• What if I draw another block on the left side and we assume that the middle block has friction. Can I use this method too? Is the force of tension the same? How can i deal with the force of friction?
• the big part that I know is that the tension on the left side is the same through out that side but the right side could have a completely different tension or it could be the same it depends on the mass of the new block
(1 vote)
• so when i say m/s squared some people tend to think meters per second per second so that is why i am confused is meters/s squared another way of saying meters per second per second?
• Yes, because in algebra (m/s)/s is m/s^2
• A bird is sitting in a large closed cage which is placed on a spring balance. The spring balance records a weight of 25 N. The bird (mass = 0.5kg) flies upward in the cage
with an acceleration of . The spring balance will now record a weight of
(a) 24 N (b) 25 N (c) 26 N (d) 27 N
Although this question isn't related to tension, it definitely has something to do with systems, internal forces and external forces. The correct answer given is (B) but if the bird isn't sitting on the bottom of the cage anymore, shouldn't the weight recorded by the spring balance reduce?