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Find min/max accel. for block to stay on wedge (static fric)

  • #1
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This HW problem due date has already passed. I had no problem with part a, but struggled thinking about parts b and c. I saw the solutions for parts b and c, but still don't exactly get it. I'll state the questions below, and tell you my way of interpreting the solution now. Could you tell me where my thinking goes wrong?

1. Homework Statement

A block rests on a wedge inclined at angle θ (this angle is with respect to the horizontal). The coefficient of friction between the block and plane is μ. a) Find the maximum value of θ for the block to remain motionless on the wedge when the wedge is fixed in position. b) The wedge is given a horizontal acceleration a. Assuming that tan θ > μ, find the minimum acceleration for the block to remain on the wedge without sliding. c) Repeat part (b), but find the maximum value of the acceleration.

2. My attempts/interpretations of the solution: Please correct me where I'm wrong

Part A: It is best to attack part a with a rotated coordinate system where x' may become our parallel axis and y' our perpendicular axis. As we increase theta, we find that our normal force decreases in magnitude, weakening the magnitude of the force of static friction that depends on the normal force magnitude. If we're concerned with finding how far we can increase this angle while allowing the block to stay at rest, we'll looking for the point where the static friction force reaches its maximum value and still can just counter the downward, parallel component of gravity.

Carrying out the appropriate math, we obtain tan(theta_max) = μ, resulting in theta_max = arctan(μ)

Parts B and C: Okay, so we're told that tan(theta) > μ, meaning that theta > arctan(μ) and that theta > theta_max: we're exceeded our angle limit and this block is going to slide down if nothing is done. With the force of static friction only depending on the the coefficient of static friction, μ, which is not going to change, and the magnitude of our normal force, our only hope to keep the block stationary is to accelerate the wedge rightwards (the diagram is oriented such that the right-angle of the wedge makes a correctly oriented letter "L") so that an increased normal force may be supplied to the block, increasing the magnitude of the static friction force accordingly.

In this case though, we are tasked with finding the minimum and maximum accelerations to be applied to the wedge for this block at such an angle to stay on the edge without sliding. The component of the normal force in the x-direction acting on the block will allow for it to accelerate uniformly with the wedge, so we may use a_min and a_max in our equations for the blocks x and y motion.

Looking at the answers, I see that the minimum acceleration case illustrates the force of static friction going upwards, parallel to the incline. Well, okay. We know that we must accelerate this system so that the normal force is of a greater magnitude so that the force of static friction is strong enough to prevent this block from slipping. Well, to counter the downwards, parallel component of gravity, it makes sense to have the force of static friction point up the plane. Now we must write out our equations for the net forces acting on the block in the x & y directions (with x-comp of N being Nsin(theta) and the y-comp being Ncos(theta)), acknowledging that the y-acceleration must be zero. We also take fstatic to = fstatic_max in our calculations, as we're already at an angle that requires this to be true. The math seems straightforward for there: a_min = g*([tanθ-μ]/[1+μtanθ]).

However, the maximum acceleration case illustrates the force of static friction going downwards, parallel to the incline. Why is this the case? I just don't get the intuition for that case. So if the force of static friction were to point downwards, it would have an x-component contributing to the rightwards acceleration of the block. If this were to happen, will a now be maxed (rather than with the f_static_x comp going opposite the direction of acceleration a)? Because we do the rest of our math assuming that the vertical acceleration of the block is zero (Ncosθ = f_static_vertical + mg_vertical), does this allow for us to now accept that the acceleration is maxed (with f_static now even contributing to that rightwards acceleration) and that the block is not slipping (because there is no vertical acceleration)?

Is this because, no matter what acceleration a we give the system, it can always be higher and reach its a_max if we allow the x-component of static friction to be in the same direction of that acceleration, meaning that we make f_static downwards/parallel (and a similar logic for the minimum acceleration)? After taking care of that, is the non-slipping condition then handled by just forcing the vertical acceleration of the block to be zero (i.e., net_force_vert equals zero)?
 
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Answers and Replies

  • #2
haruspex
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our only hope to keep the block stationary is to accelerate the wedge rightwards ... so that an increased normal force may be supplied to the block, increasing the magnitude of the static friction force accordingly.
There's a bit more to it. Without the wedge's acceleration, the block's acceleration would have a horizontal component. Accelerating the wedge allows the block to have some horizontal acceleration without slipping, so the frictional force does not need to be as great.
Specifically, Ffric=m(g sin(θ) - a cos(θ)).
the maximum acceleration case illustrates the force of static friction going downwards, parallel to the incline. Why is this the case?
Because if the wedge accelerates too fast the block may slide up the wedge, and friction acts to oppose relative motion of the surfaces in contact.
 

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