Masses connected by a pulley on a frictionless surfaceby DLH112 Tags: connected, frictionless, masses, pulley, surface 

#1
Oct1612, 01:58 PM

P: 20

1. The problem statement, all variables and given/known data
A block of mass 2.6 kg (M1) lies on a frictionless surface. Its connected by a massless pulley and string hanging over the edge of the surface of 3.17 kg (M2). Calculate the acceleration of M1(part 1). calculate the tension in the cord(part 2). the cord is massless/weightless/unstretchable, theres no friction or rotational inertia in the pulley. 2. Relevant equations F=ma 3. The attempt at a solution I think the pulley transfers the vertical Fg 9.8 x 3.17 to be horizontally pulling on M1. What I'm mostly unsure of is how/if gravity/normal force/the mass of m1 affects it without friction. 



#2
Oct1612, 09:29 PM

P: 20

A wrong solution I got was that the 31.066 (9.8 x 3.17) N were pulling the 2.6 Kg block horizontally, so the answer would be a = 31.066/2.6 . or a = 11.948 m/s^2 . What am I missing?
I've solved this now. I didn't realize how the mass on the frictionless surface factored into the problem. 



#3
Oct1612, 10:17 PM

P: 46

You still need to factor in the mass of the block off the table. Remember, you're trying to find the acceleration of the system.




#4
Oct1612, 10:28 PM

P: 2

Masses connected by a pulley on a frictionless surface
I'm having the same problem. Can someone please help. Thankss




#5
Oct2312, 03:19 AM

P: 37

I am having similar troubles solving a problem exactly like this. the only difference is the masses.
When the Free Body diagram is drawn, the mass of the hanging object pulls the object on the table to the right, so the force of the hanging object effects the right side of the diagram? [itex]\uparrow[/itex] Fn ° [itex]\rightarrow[/itex] hanging mass [itex]\downarrow[/itex] Fg 



#6
Oct2312, 09:21 AM

P: 46

That's exactly right. So what's the problem?




#7
Oct2312, 09:27 AM

P: 37

i don't know how to solve for the acceleration with only the mass of the 2 objects...




#8
Oct2312, 09:31 AM

P: 46

I'm not sure what you mean by "only the mass of the two objects" but it's just Newton's second law. You need to add up the masses when solving for the acceleration.




#9
Oct2312, 09:40 AM

P: 37

Can you show me the set up?




#10
Oct2312, 12:18 PM

P: 46

I'm not sure how to draw on here, but I'll try to explain it to you the best way I can.
If you draw out what's happening here, it's easier to understand. You have two masses connected by a string. One of those is hanging off the edge and one is sitting on a table, or whatever it's sitting on. There is only one force contributing to the movement of both these blocks, and that is gravity. Keep in mind that it's only the block hanging off the side of the table that gravity affects. By looking at a diagram of this situation though, it should be obvious that this block will not accelerate at 9.8 m/s/s. Why? Because the block on the table has inertia. Its mass is stopping the hanging mass from acclelerating at 9.8. So you should know Newton's second law. And keep in mind that both objects will accelerate at the same rate because the rope will keep the two blocks the same distance from each other unless the rope stretches. But it doesn't, so don't worry about that. They accelerate at the same rate, so if you know the mass of the system, which is the two masses added together, and you know the force acting on the system, which is just the weight of the block hanging off the table, then you know the acceleration of the system. Hope this helps. 


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