Recent content by fl00f

That's a new concept I wasn't familiar with. I'll read more into it. Thanks so much for your help! You're very good at explaining things :)

Ohh hmm! I solved the question using your suggestions: 1. m3*a = m2*g a = m2*g / m3 2. F applied = (m1 + m2 + m3) a = (m1 + m2 + m3) * (m2*g / m3) = 363 N which is the correct answer. It now makes sense to me that the force that needs to be resisted is the downward force of m2*g and that...

The inertial force of block 3 which acts leftward and is equal in magnitude to the force causing the system to move rightward

If by block 1 you mean m3 in the image then the tension would be equal to (m2)g in magnitude So it would be erroneous to consider the force of gravity acting on the system to equal (m3+m2)g, b/c it would only be (m2)g, as m3 does not act in the vertical direction?

When the system is pulled to the right, m3 resists (I presume that this is the inertial force? ..I'm not quite sure of what inertial forces are). If this resisting force is equal to the force of gravity pulling m2 and m3 downwards [(m2+m3)g] then m2 will not fall? However, I don't believe I...

In drawing a FBD, I see that the Force applied is at 90° to the force of gravity pulling m2 down. How do perpendicular forces cancel each other? Or is that not the case...

The particular situation is below: "For the system shown in the attached image, find the force required to prevent the mass m2 from descending. Use the values of m1 = 5.25 kg, m2 = 6.50 kg and m3 = 2.500 kg. Assume that all surface are frictionless and that the ropes do not stretch. (Also, in...