How Does Friction Affect the Motion of Three Blocks?

  • Thread starter Thread starter complexc25
  • Start date Start date
  • Tags Tags
    Blocks Friction
Click For Summary
SUMMARY

The discussion focuses on the application of Newton's second law to a system of three blocks affected by kinetic friction. The coefficient of kinetic friction is 0.560, with masses m1 = 0.150 kg and m2 = 0.250 kg. The user initially miscalculated the mass m3 required for the system to move at constant speed, incorrectly equating tensions and weights. The correct approach involves applying separate equations for each mass and recognizing that the tensions T1 and T2 are not equal.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with the concept of kinetic friction
  • Basic knowledge of tension in a system of connected masses
  • Ability to set up and solve simultaneous equations
NEXT STEPS
  • Learn how to apply Newton's second law to multiple objects in motion
  • Study the effects of friction on motion in physics
  • Explore the concept of tension in pulley systems
  • Practice solving problems involving connected masses and friction
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of applying Newton's laws in practical scenarios.

complexc25
Messages
12
Reaction score
0
You are given that the coefficient of kinetic friction between the block and the table in the figure is 0.560, and m1 = 0.150 kg and m2 = 0.250 kg.
(a) What should m3 be if the system is to move with a constant speed?

this is what i have:
T2 - T1 - f = m3a

m2g - m1g - ukg = m3
2.45 - 9.65 - 5.488 = 4.508kg

the answer is wrong, what is it that i am doing wrong?
 

Attachments

  • 4-36.gif
    4-36.gif
    7.3 KB · Views: 487
Last edited:
Physics news on Phys.org
complexc25 said:
this is what i have:
T2 - T1 - f = m3a
This is good.

m2g - m1g - ukg = m3
2.45 - 9.65 - 5.488 = 4.508kg
This is not good. For one thing, the tensions are not equal to the weights.

Hint: Apply Newton's 2nd law to each mass separately. You took care of M3, now come up with equations for M1 & M2.

You'll combine all three equations to solve for the acceleration and the tensions.
 
i don't understand, i have for:
m1
T - m1g = m1a
m2
m2g - T = m2a
 
complexc25 said:
i don't understand, i have for:
m1
T - m1g = m1a
m2
m2g - T = m2a
Looks good, except that the tensions are different. Use T1 for m1 and T2 for m2.
 

Similar threads

Accelerating pulleys (3 pulleys and 3 masses)
  • · Replies 5 ·
Replies
5
Views
2K
How much force does the 20 kg block exert on the 10 kg block?
  • · Replies 17 ·
Replies
17
Views
2K
Friction, Mass and Acceleration: Analyzing Block Motion
  • · Replies 6 ·
Replies
6
Views
2K
Two blocks, a fixed pulley and friction
  • · Replies 4 ·
Replies
4
Views
3K
Block-pulley system and rotational dynamics
  • · Replies 5 ·
Replies
5
Views
5K
How Does Friction and Acceleration Affect Tension in a Three-Block System?
  • · Replies 18 ·
Replies
18
Views
2K
Friction in a 3 block and pulley system
  • · Replies 4 ·
Replies
4
Views
4K
Angular Acceleration and Linear Acceleration of a Pulley
  • · Replies 4 ·
Replies
4
Views
2K
Calculating tensions and acceleration
  • · Replies 10 ·
Replies
10
Views
3K
Incline with friction and two blocks
  • · Replies 5 ·
Replies
5
Views
3K