Solving Newton Law Problem: Find Acceleration & Tension

AI Thread Summary
To solve the problem of finding acceleration and tension in a system of four masses on an incline, first identify all external forces acting on the system, including the gravitational components parallel to the incline and the applied force. The correct formula for acceleration is derived from the net force divided by the total mass, taking into account both the applied force and the gravitational force component. When calculating tensions, apply Newton's second law (F=ma) to each mass individually, ensuring to include the tension from connected strings and the weight of each mass. It's crucial to treat the masses as a single system for acceleration but analyze them separately for tension. Understanding these principles clarifies the confusion around the problem.
moooocow
Messages
12
Reaction score
0
It is almost time for the end of term exam and I am looking over my old exams and am confused by one problem. Masses m1,m2,m3,m4 are cnnected together(by a "string") on an incline theta = 22 degrees. m1 and tension = T1(connects m1 and m2) start nearest the bottem of the ramp, and a force F = 30N is pulling on m4 parallel to the ramp. We had to find the acceleration of m2 and the m3 tensions. The acceleration of course is the same for all of them.

I got a = F costheta / (m4 + m3 + m2 + m1) This got marked wrong, if anyone knows why I would greatly appreciate it. Thank you
 
Last edited:
Physics news on Phys.org
Hint: the force F is parallel to the ramp.
 
I am just really confused on this one, is it not needed to take the force in the "x" direction because since the acceleration and force are on the same "line", which we could use as the "x" axis? When finding the tension I had, for one example, m4*a = F cos(theta) - T3 cos(theta) and this was wrong too. I am really confused here if anyone could help with this type of problem, thank you very much.
 
I guess my "hint" wasn't that helpful. :smile: I'll try again.

To find the overall acceleration, first find all the external forces acting on the masses (taking them as a system), taking their components parallel to the plane. The weight acts down, so you need to find the component of the weights acting down the plane. (Careful with the angles!) The only other force is the applied force F. Now add these forces, then apply
Fnet = mtotal a

To find the tension in the strings, apply F=ma to each mass separately and solve the set of equations.

Edit: Just to be clear, the first comment is a "short cut". By treating the masses as a single system, you can ignore the tensions in the strings--since they are all internal forces. If this is confusing you, forget it. Just apply F=ma to each mass separately: you'll get the same answer.
 
Last edited:
So the force on the individual pieces, and the object as a whole would just be the applied force to the x plane and the gravitational force to the x plane?

a = F + mg sin(Theta) / m

Thank you very much for the help, I have no idea why I was having so much trouble.
 
Last edited:
Originally posted by moooocow
So the force on the individual pieces, and the object as a whole would just be the applied force to the x plane and the gravitational force to the x plane?

a = F + mg sin(Theta) / m
Right! Treating the four masses as a single object, that's how you can find the acceleration. (I assume you meant a = (F + mg sin(Theta)) / m )

When treating each mass separately (which you must do to find the tension each string), don't forget to include the tension in any string touching the piece as well as the weight.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top