Two Masses One on Table, One Hanging

  • Thread starter Thread starter freeofwork
  • Start date Start date
  • Tags Tags
    Table Two masses
AI Thread Summary
The discussion revolves around a physics problem involving two masses, one on a table and one hanging, connected by a string. The key equations used include F=ma and the frictional force equation, with a coefficient of friction of 3.00 between the block and the table. Participants clarify that the tension in the string is influenced by both masses, and both blocks will accelerate at the same rate due to the inextensibility of the string. Newton's third law is applied to explain the interaction between the two blocks, emphasizing that the hanging block exerts a pull on the block on the table and vice versa. Overall, the focus is on understanding the forces acting on each mass and their resulting acceleration.
freeofwork
Messages
44
Reaction score
0

Homework Statement



A block of mass1 5.0 kg is on a table that bonds a coefficient of friction of 3.00. Mass1 is attached to Mass2 by a string, and Mass2 is hanging over the table. The system is released from rest for it to accelerate.

Homework Equations



F=ma
F fric = μFnormal

The Attempt at a Solution



\sum Fx of Mass1 = F tension - F friction

\sum Fy of Mass1 = 0

\sum Fx of Mass2 = 0

\sum Fy of Mass2 = FW - F tension

Is tension created by the mass1 or mass2?
im guessing both? but that doesn't make any sense...
 
Physics news on Phys.org
1. Where is m3?
2. Is coefficient of friction of 3.00 is between the block and table?
 
azizlwl said:
1. Where is m3?
2. Is coefficient of friction of 3.00 is between the block and table?

no m3...

2. yes
 
ok wrong interpretation.

Find net force on each block.
Both will accelerate at equal rate for the string is massless and inextensible.
 
Last edited:
oops one on table, one hanging
 
Use Newton's 3rd law.
The hanging block will pull the block on the table.
The block on the table will pull the hanging block with equal force.
 
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 '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