I'm trying to figure out whether ug = a from the argument F_friction = uF_normal = u(mg) = m(A_friction) -> ug = a is sensible.
where:
u (coeff. of friction) is given for a mass of 'm' traveling on an even plane on Earth where the gravitational constant is 'g'.
F_friction = uF_normal = umg
F_friction = umg = ma; masses cancel therefore ug = a
Setting F_friction = ma seems like a dubious step since Net_(F)=ma; Newton's second law is generally reserved for the motion of a mass 'm' undergoing an acceleration 'a' as a result of the net force. Force of (moving) friction is generally not a stand-alone force so i'm kind of skeptical. This equation seems to imply that if Force of friction is the net force ... is that possible? By what reasoning is this equation justified (if it is)?
where:
u (coeff. of friction) is given for a mass of 'm' traveling on an even plane on Earth where the gravitational constant is 'g'.
F_friction = uF_normal = umg
F_friction = umg = ma; masses cancel therefore ug = a
Setting F_friction = ma seems like a dubious step since Net_(F)=ma; Newton's second law is generally reserved for the motion of a mass 'm' undergoing an acceleration 'a' as a result of the net force. Force of (moving) friction is generally not a stand-alone force so i'm kind of skeptical. This equation seems to imply that if Force of friction is the net force ... is that possible? By what reasoning is this equation justified (if it is)?
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You have not considered the direction of motion.
Considering the direction of motion as positive, the frictional force is negative
and hence F_(friction) = - μmg .
You are confusing yourself with the net force and applied force.
Consider a force F is applied on an object.
And let the frictional force = - μmg .
The net force is F – F( friction) = (F – μmg)
The object will have an acceleration a = (F – μmg) / m = (F/m – μg).
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If the applied force is also = μmg
The net force is = (F – μmg) = (μmg – μmg) = 0
It implied there is no acceleration which implies that the object may either at rest or move with constant velocity. ( note not with constant acceleration)
The net force acting on the object is zero.
======================================…
If ,as you have stated, ma = μmg, it implies from
ma = (F – μmg) that μmg = (F – μmg)
F = 2 μmg
That is the applied force is twice the frictional force in magnitude and the net force is (ma or μmg)
--------------------------------------…
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Considering the direction of motion as positive, the frictional force is negative
and hence F_(friction) = - μmg .
You are confusing yourself with the net force and applied force.
Consider a force F is applied on an object.
And let the frictional force = - μmg .
The net force is F – F( friction) = (F – μmg)
The object will have an acceleration a = (F – μmg) / m = (F/m – μg).
--------------------------------------…
If the applied force is also = μmg
The net force is = (F – μmg) = (μmg – μmg) = 0
It implied there is no acceleration which implies that the object may either at rest or move with constant velocity. ( note not with constant acceleration)
The net force acting on the object is zero.
======================================…
If ,as you have stated, ma = μmg, it implies from
ma = (F – μmg) that μmg = (F – μmg)
F = 2 μmg
That is the applied force is twice the frictional force in magnitude and the net force is (ma or μmg)
--------------------------------------…
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There is another option, in addition to those mentioned above, that might apply to your situation. Your object may be moving at some constant velocity (zero net force) when it all of a sudden encounters a surface with friction. There will then be a net force on the object due only to friction and your eq. would be;
ug = -a (minus because friction force opposes the motion)
So your object would then decelerate until it came to rest, at which time the friction force goes to zero because the friction force is only present if there is motion or impending motion.
ug = -a (minus because friction force opposes the motion)
So your object would then decelerate until it came to rest, at which time the friction force goes to zero because the friction force is only present if there is motion or impending motion.