Force on m by 2m such that m is stationary on M

  • Thread starter Thread starter Cali37
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The discussion revolves around a physics problem involving two masses, m1 and M, connected by a massless string, with a focus on static friction. Participants are tasked with determining if a coefficient of static friction μs exists that allows m1 to remain stationary on M. The problem also explores whether the scenario changes when the masses are interchanged. Users express confusion over missing information necessary to solve the equations provided. Clarification on the problem setup and additional details are requested to proceed with the calculations.
Cali37
Messages
3
Reaction score
0

Homework Statement


A mass m1 is placed upon a second mass M on a horizontal surface. A massless string connects mass m1 to mass m2. If the coefficient of static friction between M and the horizontal surface is μs, answer the following:
a) There is a coefficient of static friction μs between M and m1 such that m1 will remain stationary on M. Prove or disprove.
b) If the masses are now interchanged such that m1 is switched with m2, then the answer to part (a) remains the same. Prove or disprove.

Homework Equations


m1+M=m2
ƩFx=F-T
ƩFy=N-mg


The Attempt at a Solution


N=mg
 
Physics news on Phys.org
Please supply the rest of the information. There's obviously some missing.
 
Thats all that my teacher gave me, you can see why I am struggling. Thanks anyway :)
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

A cylinder connected to a hanging mass
Replies
21
Views
274
Required force for sphere start rolling
Replies
2
Views
675
Normal force at the base of a ladder
Replies
20
Views
2K
Final Velocity of a car with a opposing decreasing drag force
2
Replies
57
Views
2K
Mechanical Physics: Coefficient of Friction
Replies
20
Views
3K
How can I find the critical mu of friction?
Replies
32
Views
22K
Net Force on 2m Mass: Prove/Disprove?
Replies
1
Views
2K
Newton's law problem: Pushing 2 stacked blocks on a horizontal table
Replies
13
Views
3K
Work done during a collision -- Change in Kinetic Energy & change in Momentum
Replies
1
Views
2K
Determine the friction force in a stacked block scenario
Replies
23
Views
3K

Hot Threads

Recent Insights

Back
Top