Friction on a black on an inclined plane

Click For Summary
SUMMARY

The discussion focuses on determining the motion of a 2 kg block on an inclined plane with a static friction coefficient of 0.5 at a 20-degree angle. The calculations reveal that the frictional force is 0.47 N, while the net force acting on the block results in an acceleration of 3.19 m/s². However, upon reevaluation, it is concluded that the block does not move when the angle is 20 degrees, as the net force is negative, indicating that the static friction is sufficient to prevent motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its coefficient
  • Basic trigonometry for calculating forces on inclined planes
  • Familiarity with free body diagrams
NEXT STEPS
  • Study the concept of static versus kinetic friction in detail
  • Learn how to draw and analyze free body diagrams for inclined planes
  • Explore the effects of varying angles on the motion of objects on inclined planes
  • Investigate the role of mass and friction in determining acceleration
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of static friction and inclined plane problems.

sara_87
Messages
748
Reaction score
0

Homework Statement



A block of mass 2kg is on an inclined plane and held steady. The coefficient of static friction u between the block and the plane is 0.5. Determine whether or not the block moves when released if the angle the plane makes with the horizontal is
(a) 20 degrees


The Attempt at a Solution



to find n i did this:

2*10*cos20 = 0.94

so friction is 0.47

then i did this:

2*10*sin20 - 0.47 = 6.37

so acceleration is 6.37/2 = 3.19m/s^2

so it looks like the block does move when tilted by 20 degrees...

but i feel like i got quite a big number for the acceleration, am i doing something wrong? am i supposed to approach this question differently?
 
Last edited:
Physics news on Phys.org
If the downward force is F and the frictional force is F_R

and [itex]F=mgsin\theta[/itex] and [itex]F_R=\mu mgcos\theta[/itex]

consider what happens if [itex]F-F_R<0[/itex] and what happens if [itex]F-F_R>0[/itex]

one says that there will not me motion and the other says that there will be

sara_87 said:
2*10*cos20 = 0.94

so friction is 0.47

then i did this:

2*10*sin20 - 0.47 = 6.37
Are you sure that is correct? I mean the frictional force part
 
Last edited:
sorry made a stupid mistake:

2*10*cos20 = 0.94

and i found a more reasonable way of doing it

the answer is supposed to be negative so the the box doesn't move
 

Similar threads

Replies
2
Views
898
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K