Block sliding down wedge max value problem

Click For Summary

Homework Help Overview

The discussion revolves around a block resting on an inclined triangular wedge, focusing on the conditions under which the block remains motionless. The problem involves analyzing forces related to friction and acceleration, with specific attention to the angle of the wedge and the coefficient of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the balance of forces acting on the block, including gravitational components and friction. There are attempts to derive conditions for the block to remain stationary on the wedge, considering both fixed and accelerating scenarios. Questions arise regarding the direction of static friction when the block is subjected to varying accelerations.

Discussion Status

The discussion is active, with participants exploring different aspects of the problem. Some have provided insights into the forces at play and the role of static friction, while others express uncertainty about specific conditions and the implications of friction direction. No consensus has been reached regarding the direction of friction under certain conditions.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. The problem involves assumptions about the wedge's fixed position and the nature of friction in different scenarios.

bodensee9
Messages
166
Reaction score
0
Hello:


I have an inclined triangular wedge of angle Ø (meaning, this wedge makes angle Ø with the horizontal) and a block is lying on the wedge. the coefficient of friction between the wedge and block is u.

(a) determine the max value of Ø for the block to remain motionless on the wedge when the wedge is fixed in position.
Let m = mass of block, then
This part is just umgcosØ = mgsinØ. So then tanØ = u, and u = arctanØ

(b) the wedge is given horizontal acceleration a. Assuming that tanØ < u, find the maximum and minimum acceleration so that the block doesn't slide. I think for the minimum, I will just set friction along the wedge = the component of acceleration along the wedge + the component of gravity on the block along the wedge. But I'm not sure what to do about finding the maximum acceleration? I think not sliding means that they are both accelerating horizontally at a.

Thanks.
 
Physics news on Phys.org
For the maximum, they mean not accelerating up the incline.

Remember that friction is bi-directional. So for the maximum, you need to overcome via acceleration, both the downward component of gravity along the incline as well as the friction resisting any motion.
 
A good way I've found to approach these problems is to think in terms of how much force the static friction can enact on the block.
Remember that fs max = µs * N = µs*mg*cosΘ

Also, not sliding would mean that the block is at rest relative to the wedge. If you know how to work with fictitious forces, I suggest you draw a free body diagram of the block, from the accelerated frame of the block, with fictitious forces, and say that the block is at rest and see where that gets you.

A good point to make here (And please, someone, correct me if I'm wrong) is that you should also take note of two cases. One where without friction the block would slide down the slope, and one where without friction the block would slide UP the slope. Remember that fs is a vector, and its direction is opposite that of the expected velocity.
 
oh right, that helps. i think the difference is that friction points in opposite directions. thanks.
 
Hi,

Even though I know how to 'solve' this problem mathematically, something about it still bothers me, physically speaking. So having calculated the minimum and maximum accelerations (lets call them amin and amax), suppose we give the block an acceleration a strictly between these two values. i know the block won't move relative to the wedge. But how can we know the direction of (static) friction (upwards or downwards)?

Thanks
 

Similar threads

  • · Replies 37 ·
2
Replies
37
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 11 ·
Replies
11
Views
7K