enter image description here

An "object" that has the shape of an inclined plane has a slot that carries a roller bearing attached to an extension spring. When a force F1 is applied to the left, the object moves and the spring stretches (position A). When F1 is released, the object moves to the right due to the restoring force of the spring and rolling contact (position B). It is assumed that all surfaces are frictionless and the object can only slide left-right. It is constrained to move or rotate in any other direction. The spring is also assumed to deform in the y-axis only. When I draw a free body diagram of the forces (Fn being the normal force) to find the F1 to keep the object at position A, I get the equation as shown (by resolving components on the inclined plane). This shows both forces acting in the same direction although there is a force (a component of Fs1) that moves the object to the right. Would someone please clarify this concept? Why the two forces appear to be added and not subtracted to get the net force equal to zero. I am really confused.


1 Answer 1


I think you are confusing which forces act on which object, but to be fair, I do not understand your solution at all. It seems to me to be wrong (but maybe I am just not understanding it).

You are asking about balance for inclined plane, therefore there needs to be only forces which act on the inclined plane in your free body diagram. I have no idea how did you get the forces you have there.

Your problem states that the inclined plane can move only horizontally, this means, that there is vertical force of constraint that compensates all vertical components of all other forces acting on the inclined plane. This force is missing from your free body diagram.

Next thing is that you do not wish to decompose the forces in the direction of the inclined plane. You are not interested in motion in this direction, you are interested in left-right motion, which is the only one allowed. Therefore, your condition is that the horizontal component of net force acting on the inclined plane should be zero (again, vertical component of net force is zero due to the force of constraint, which keeps the motion only horizontal, i.e. left-right)

Another thing is, that because the surfaces are frictionless, the only possible interaction between roller bearing and an inclined plane is by normal forces. Your $F_{s1}$ is however not normal to the inclined plane and thus this cannot be the force with which roller bearing acts on the inclined plane.

The free body diagram for an inclined plane should look like this (ignoring gravitational force acting on the inclined plane, but this is vertical so it is irrelevant for our problem. Let us say it is included in the force of constraint):

enter image description here

And the equation for vertical components is $$F_1=F_n\sin(\varphi)$$

On the roller bearing we have this freebody diagram:

enter image description here

where $F_{react}$ is the reaction force to the normal force acting on the incline plane and $F_{constr}$ is again force of constraint, that keeps the string vertical. Thus: $$F_{s1}=F_{react}\cos(\varphi)=F_n \cos(\varphi)$$ And combining the equations we get $$F_{s1}=\frac{F_1}{\tan\varphi}.$$

  • $\begingroup$ Thank you for the response. What I considered was that the roller exerts a force equal to the Fs or Fs1 in the upward direction on the top inclined plane. As you can see roller is sandwiched between two surfaces but free to move. The normal force then acts in the downward direction. I still don't understand how you take the normal in the upward direction. I also don't understand the horizontal Fcoonst in your second FBD. Would you please elaborate the two points. $\endgroup$
    – rayan
    Commented Jun 9, 2021 at 10:30
  • $\begingroup$ @ryan as I said, roller cannot exert force in the upward direction, only in the direction perpendicular to the the top inclined plane, because the surface is frictionless. So only the component of the upward force produced by the string on the roller which is perpendicular to the plane gets transferred to the inclined plane. This is what I called normal force. $\endgroup$
    – Umaxo
    Commented Jun 9, 2021 at 10:39
  • $\begingroup$ @ryan if there would be no constraint on the string to remain vertical, there would be net horizontal force acting on the roller and the string would start to rotate or bend. This is however forbidden so horizontal Fconst is force that compensates this. $\endgroup$
    – Umaxo
    Commented Jun 9, 2021 at 10:42
  • $\begingroup$ @ryan your downward normal force acts on the roller not on the inclined plane, so this is what I call $F_{react}$. $\endgroup$
    – Umaxo
    Commented Jun 9, 2021 at 10:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.