# Angle of the slope when the object start moving

## Homework Statement

when the object is moving , why the theta is larger than the theta(s) when the object is started moving ?
I was told that when the object is moving , the Fk will become smaller than Fs , right ? so, IMO , the theta should be larger than the theta(s) , am i right ?

## The Attempt at a Solution

tan(theta k)= Fk / N = = μ k (N) / N
tan (theta s) =Fs / N = μ s (N ) / N
angle is directly proportional to μ k and μ s

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mfb
Mentor
when the object is moving , why the theta is larger than the theta(s) when the object is started moving ?
They could have written >=, but the exact border case rarely has physical relevance - no slope is perfectly flat and so on.
I was told that when the object is moving , the Fk will become smaller than Fs , right ?
Sliding friction is usually smaller than static friction. Once the object starts moving, you can reduce the tilt angle and the object will continue sliding.

They could have written >=, but the exact border case rarely has physical relevance - no slope is perfectly flat and so on.
Sliding friction is usually smaller than static friction. Once the object starts moving, you can reduce the tilt angle and the object will continue sliding.
do u mean when the object is sliding , no matter the angle is larger than μ s or the angle is slightly smaller than μ s , the object will still sliding down the plane ?

mfb
Mentor
μ s is not an angle.

There is an angle ##\theta_0## where the object starts sliding (it starts sliding for all angles larger than that, and does not start for all angles smaller than that). There is a different angle ##\theta_1 < \theta_0##, between those two angles an object that is sliding keeps sliding (forever), but won't start sliding on its own if it is at rest.

θ1<θ0
as you stated , why the book gave θ(when it's moving ) will bigger than θ(when it's about to move ) ?

mfb
Mentor
The logic is in the other direction: if the angle is larger, it is certainly moving.

• goldfish9776