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Imagine a seesaw with blocks of mass m each placed on extreme end. The axis of rotation is in center. The length of seesaw is L.

If we are given a position of seesaw in which the seesaw is making 30 degree angle with horizontal. Will it be called equilibrium ?

Because when I see the equilibrium equations

Sum Forces in x direction: 0 (because no force in x direction)

Sum Forces in y direction: N - mg - mg = 0

Sum of torque: mgLsin(30)/2 - mgLsin(30)/2 = 0

But it very hard to believe that with equal weights at equal position from center will be at equilibrium even in inclined position.

I am not very strong in physics can someone help me understand what is wrong here

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It depends on the shapes of the two equal blocks, or, more specifically, on the location of their centres of mass.

If a line joining the centres of mass of the two blocks runs through the pivot then the seesaw will indeed be in equilibrium at any angle of inclination - the net moment about the pivot will always be zero. This assumes the blocks stay in position, and do not slide along the seesaw as it tips.

However, if a line joining the centres of mass of the two blocks is above the pivot then the seesaw will only be in equilibrium when it is horizontal, and this is an unstable equilibrium. At any inclination away from the horizontal there will a moment that tends to tip the seesaw further away from the horizontal. Intuitively, you can see that the centre of mass of the whole system is lowered if it tips in either direction, so it is energetically favourable for the seesaw to tip further - it is like a pencil balanced on its point.

On the other hand, if a line joining the centres of mass of the two blocks is below the pivot then the seesaw will also only be in equilibrium when it is horizontal, but this time this is a stable equilibrium. At any inclination away from the horizontal there will a moment that tends to tip the seesaw back towards the horizontal. You can see that the centre of mass of the whole system is raised if it tips in either direction, so it is energetically favourable for the seesaw to return to the horizontal.

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  • $\begingroup$ Could you possibly elaborate further how it depends on the line connecting the centers of mass of each block? I've been searching this topic for days and still can't understand how a torque imbalance would even be possible if one set the 2 equal masses on an incline at equal distances from the fulcrum $\endgroup$
    – MattGeo
    Commented May 31, 2022 at 13:29

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