A heavy box refuses to slide as Imre raises his force on it. Then, the instant it breaks free, it jerks forward, faster than he meant.
Every mover knows that lurch.
The lurch comes from two coefficients.
For most surface pairs, the maximum static friction is larger than the kinetic friction: μ_s > μ_k.
Starting a slide takes more force than keeping it going.
For most surface pairs, which friction force is larger?
Give the box numbers. Its maximum static friction is 120 N. Once it is sliding, the kinetic friction is only 90 N.
To start the slide, Imre's force has to pass 120 N.
The instant sliding begins, the friction drops from 120 N to 90 N, while his force still sits just past 120 N.
Tilda says: 'Just as a dresser breaks free and starts to slide, the friction on it drops.' Is she correct?
Suddenly the box has a net forward force of about 30 N where a moment before it had none. The box jerks forward. That is the lurch.
So objects lurch the moment they break free: the friction falls to the smaller kinetic value faster than the mover can reduce the applied force.
A heavy chest lurches forward the instant it starts to slide. Which statement explains the lurch?
'Most pairs' is honest. For a few material pairs the two coefficients are nearly equal, and the lurch is barely felt.
For a filing cabinet on tile, the maximum static friction is 200 N and the kinetic friction is 160 N. Rowan's applied force reaches 200 N just as the cabinet breaks free, and for a moment it stays there. Calculate the net force on the cabinet just after it starts to slide.
Write down the values in the question:
applied force = 200 N
kinetic friction = 160 N, against the slide
net force = ?
Write down the equation:
net force = applied force − kinetic friction
Substitute in the values:
net force = 200 N − 160 N
Simplify:
net force = 40 N, forward
The net force on the cabinet just after breakaway is 40 N, forward.
Why does a heavy box jerk forward just as it starts to slide?
Before the slide, static friction matched the applied force, so the net force was zero.
At breakaway, the friction drops from the static maximum μ_s N to the smaller kinetic value μ_k N.
The applied force has not changed, so a net forward force suddenly appears.
The friction drops to the smaller kinetic value while the applied force stays, so the box suddenly has a net forward force.
Which statement about most surface pairs is correct?
For a workbench being slid across a shop floor, the maximum static friction is 340 N and the kinetic friction is 280 N. The applied force is 340 N just as sliding begins, and for a moment it stays there. Enter the net force on the bench in N just after it breaks free.
The applied force on a heavy box rises slowly from zero until the box is sliding. Which sequence describes the friction on the box?