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I was reviewing some basic concepts in physics. I was reading a concept that says “the force needed to push/pull an object on an inclined plane (suppose friction is zero) is $$F= m\,g\,\sin(\theta)$$ My question is, if we exert the force on that object with that amount of force, wouldn’t we end up preventing the object from sliding down rather than move the object up?

Because the forces would cancel each other out and thus, no motion is made. Shouldn't the force required to move the object upwards be greater? I am thinking the force should be $F_p = m\,a + m\,g\sin(\theta)$.

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  • $\begingroup$ was that helpful ? $\endgroup$
    – Ankit
    Sep 2, 2020 at 5:51
  • $\begingroup$ Yes...it was really helpful...thanks man @Ankit $\endgroup$
    – Enrico Hendro
    Sep 2, 2020 at 8:11
  • $\begingroup$ Will do 😀 @ankit $\endgroup$
    – Enrico Hendro
    Sep 3, 2020 at 6:10
  • $\begingroup$ I have edited my answer. So look out for the last paragraph ☺️ . It will be more helpful. $\endgroup$
    – Ankit
    Sep 3, 2020 at 6:21
  • $\begingroup$ I have read the additional paragraph...it is really helpful 😊 thanks a lot man @ankit $\endgroup$
    – Enrico Hendro
    Sep 4, 2020 at 3:57

2 Answers 2

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See if the body was sliding on the inclined plane ( considering frictionless surface) with only gravitational force then it would be accelerating along the plane with an acceleration of $g\sin \theta $.

Now if you apply the same amount of force along the plane then it wouldn't come at rest but will continue its motion in the same direction but this time with a constant velocity ( which is the velocity the body had gained (since it's velocity was changing at every moment due to gravity) when the external force was applied ) .

So in order to move the body upward i.e. in opposite direction , YES you will have to apply a greater amount of force .

As @ohneVal mentioned if the body was initially going upward (because of any push ) then we just need to apply force equal to its gravitational component along the plane and the body will move upward with a constant velocity.

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  • $\begingroup$ Ah I see...I get it, that’s why when subjected with the same amount of force, an object will have zero acceleration which could also mean it is moving with constant velocity...am I getting this right?? $\endgroup$
    – Enrico Hendro
    Sep 2, 2020 at 6:07
  • $\begingroup$ @Enrico Hendro yes you are absolutely right. So first you need to stop the object by applying a greater amount of force and then with that force it can go up. $\endgroup$
    – Ankit
    Sep 2, 2020 at 6:14
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    $\begingroup$ Take into account that if the initial condition was that the block was "kicked" upwards, in other words given an initial speed upwards and immediately after we start applying the countering force to gravity, one would still obtain a constant velocity situation. The block would keep moving upwards with constant speed. So the initial condition is important $\endgroup$
    – ohneVal
    Sep 2, 2020 at 10:11
  • $\begingroup$ Thanks man 😊 @ohneval $\endgroup$
    – Enrico Hendro
    Sep 4, 2020 at 3:58
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Yes the object would then be in equilibrium. If you give it some velocity and then apply the force ($mg\sin \theta$), it would move up the slope.

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  • $\begingroup$ I see...so to summarize, as long as we are moving it with some velocity, then (mgsin(theta)) would suffice, but if we are moving it from rest (Vo at the bottom of the incline = zero) then we would need greater force than that (which is F=m.a + m.g.sin𝜃). Am I getting this right?? $\endgroup$
    – Enrico Hendro
    Sep 2, 2020 at 2:52
  • $\begingroup$ A body in physical equilibrium means ' Net external force on the body is Zero'. That also means acceleration is O. Zero acceleration means either velocity is O or body moving with constant velocity. From Physics point of view, State of Rest or State of Uniform motion are same and Body is said to be in equilibrium. $\endgroup$
    – Nitish Kumar
    Sep 2, 2020 at 5:11

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