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- Thread starter jason.bourne
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Doc Al

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Shouldn't the 'inertial force' oppose the *acceleration*, not the motion?

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hey! thanks for replying.

i was following this website. here's the link

http://lpsa.swarthmore.edu/Systems/MechTranslating/TransMechSysModel.html

it has been mentioned in this website that 'inertia force acts opposite to the direction of motion'.

did i understand it right?

mass will decelerate in the positive x direction, when it moves away from equilibrium towards +A.

it will accelerate in the negative x direction, when it is moving towards the equilibrium position.

it will start decelerating in the negative x direction, when it moves away from equilibrium position towards -A.

it will start accelerating in the positive x direction, when it moves towards equilibrium position.

i was following this website. here's the link

http://lpsa.swarthmore.edu/Systems/MechTranslating/TransMechSysModel.html

it has been mentioned in this website that 'inertia force acts opposite to the direction of motion'.

did i understand it right?

mass will decelerate in the positive x direction, when it moves away from equilibrium towards +A.

it will accelerate in the negative x direction, when it is moving towards the equilibrium position.

it will start decelerating in the negative x direction, when it moves away from equilibrium position towards -A.

it will start accelerating in the positive x direction, when it moves towards equilibrium position.

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- #4

Doc Al

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I think it's just a sloppy and misleading way to describe the direction. Note how they 'derive' the inertia force: It's just the negative of 'ma'.it has been mentioned in this website that 'inertia force acts opposite to the direction of motion'.

did i understand it right?

Since your variable x already defines the positive direction of acceleration, [itex]\ddot{x}[/itex], the inertia force is simply [itex]-m\ddot{x}[/itex]. The direction is already included: When the sign of [itex]\ddot{x}[/itex] changes, so will the sign of the inertia force.

But it's the direction of the

- #5

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Since your variable x already defines the positive direction of acceleration, x¨, the inertia force is simply −mx¨

as i defined downward x as positive, so does that mean x¨ is positive as i move downwards away from equilibrium?

i might be asking really very silly doubt but this thing is really confusing me. i want it to be sorted.

as the mass moves away from equilibrium towards +A in the downward positive x direction, if we draw a free body diagram, how do we represent acceleration?

if we take x¨, acceleration positive in the downward direction, the inertia force acts in the opposite direction i.e, upwards.

so if i write equation of motion using D'Alembert's Principle, i get:

-mx¨ - kx = 0.

but if i consider the situation where the mass is at a position, away from -A, towards equilibrium, then what is the direction of acceleration and how do i write the equation of motion?

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Doc Al

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[itex]\ddot{x}[/itex] is positive as long as the acceleration is in the positive direction, which you have defined as downward. Note that the restoring force of the spring always acts to accelerate the mass back towards the equilibrium point. So at any point below the equilibrium point, the acceleration would be upward.as i defined downward x as positive, so does that mean x¨ is positive as i move downwards away from equilibrium?

You wouldn't put the acceleration on a free body diagram, but it would act upward.i might be asking really very silly doubt but this thing is really confusing me. i want it to be sorted.

as the mass moves away from equilibrium towards +A in the downward positive x direction, if we draw a free body diagram, how do we represent acceleration?

[itex]\ddot{x}[/itex] would be negative and the inertia force would be positive. But you don't have to worry about the inertia force direction--you'd still represent that inertia force as [itex]-m\ddot{x}[/itex].if we take x¨, acceleration positive in the downward direction, the inertia force acts in the opposite direction i.e, upwards.

Right.so if i write equation of motion using D'Alembert's Principle, i get:

-mx¨ - kx = 0.

So now you are looking at the situation when the mass is above the equilibrium point. So the spring force acts downward and the acceleration is downward.but if i consider the situation where the mass is at a position, away from -A, towards equilibrium, then what is the direction of acceleration and how do i write the equation of motion?

Nonetheless, you'd still write the restoring force as -kx (since x is now negative, the force comes out positive). And you'd still write the inertial force as opposite to the acceleration: [itex]-m\ddot{x}[/itex]. (Since the acceleration will be positive, that will come out to be negative.)

So the equation of motion will be exactly the same. Realize that you don't have to know the direction of the acceleration ahead of time. Just represent the inertia force properly--with respect to your chosen coordinates. Then your equation of motion will

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if i have any doubts further, i'll get back to you.

thank you very much!

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Doc Al

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Excellent. Glad that it helped. And you are most welcome.

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