HELP Find the mass in this inclined pulley system, required to move the 2 masses

AI Thread Summary
The discussion revolves around a pulley system with two masses, m1 on an inclined plane and m2 hanging vertically. The problem involves determining whether the masses will move when released, calculating the additional mass needed on m2 to initiate movement, and finding the resulting acceleration. The calculations for forces include gravitational force, friction, and tension, leading to a conclusion that 2.8 kg is necessary to start movement and an acceleration of 5.7 m/s² once that mass is added. Participants suggest using static friction calculations and applying Newton's laws to solve for the required mass and acceleration. The focus remains on understanding the forces at play in the system to derive the correct values.
dani123
Messages
134
Reaction score
0
In this problem they show a diagram of a pulley system with an inclined plane (25 degree incline). There is a box called m1=47kg resting on the incline plane and m1=35kg hanging on the vertical line. The question reads: The coefficient of friction between m1 and the surfvace of the inclined plane are Ustatic=0.42 and Ukinetic=0.19.
a)if the masses are held in place and then released , will they start to move? ------> I know the answer is NO.
b)how much mass would you have to add to m2 to cause the masses to begin to move? ----> The answer should be 2.8kg but I don't know how the book got this answer.
c)If you said no to a) and added the mass that you calculated in b), what would be the acceleration of the masses? ---> answer given in the back of textbook is 5.7 m/s^2

ANY help with this problem would be greatly appreciated! THANK YOU SO MUCH in advance :)

For m1=47kg
Fparallel=mgsin(25)=194.7N
Fg=47*9.8=461N
Fperpendicular=Fn=417N
Ffk=uk*Fn=0.19*417=79N

Applying Newton's 2nd law to m1
Fparallel-Ff-Ft=194.7-79-Ft=47*a

For m2=35kg
Fg=35*9.8=343N

Apply Newton's 2nd law to m2
Ft-343N=35*a

combine the two equations involving acceleration and cancelling out the force of tensions... this only leads us to an equation that ends up giving us a=-2.77m/s^2... but this is where I am stuck... how do I find the mass required to get both masses to move?!

can we assume the acceleration is equal to 1 when we are trying to determine the mass of m2, since they did not specify in the question?
 
Last edited:
Physics news on Phys.org
Would it be safe to say that the acceleration is equal to 1 if they do not specify in the question?
 
You know that the force of static friction is Fs≤μmgcos(25). Calculate the static friction when a=0 and check if it is less than μmgcos(25). Then suppose that Fs =μmgcos(25) and find the mass to add to m2 so you get a=0.

ehild
 
Do I use the m2 for both of those equations?
 
The force of friction is calculated for m1=47 kg.
Do what you did in the first post: Collet the forces acting both m1 and m2, and cancel the tension.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top