What will be the normal force by the ground on the wedge?Am i doing correct please help and if not then what is wrong? Please guide.

Perhaps if your sketch was a bit neater it would be more obvious.

I can't read the mass of the block resting on the wedge.

However if the wedge is resting on the ground and the block is simply resting on the wedge then the pair of them (wedge and block) must exert a combine thrust on the ground equal to the combined mass time g ie (ma + mb)g.
Unless they then start flying, Newton's third law tells us that the ground must therefore exert a reaction equal to this.

1 hour ago, studiot said:

Perhaps if your sketch was a bit neater it would be more obvious.

I can't read the mass of the block resting on the wedge.

However if the wedge is resting on the ground and the block is simply resting on the wedge then the pair of them (wedge and block) must exert a combine thrust on the ground equal to the combined mass time g ie (ma + mb)g.
Unless they then start flying, Newton's third law tells us that the ground must therefore exert a reaction equal to this.

Sorry for the bad handwriting but what if it is accelerating like i have shown in the figure. Shouldn't the normal force by the block must have a component in the downward direction such that the wedge and the block both apply a combined "thrust". Will there be any difference in normal force by ground and if so what will it be?

Is the answer {m[block]cos^2(theta)+m[wedge]}g

If there is relative motion between the block and the wedge then you must apply Newton's Laws resolved parallel and perpendicular to the interface.

It is not then an equilibrium question.

Is this homework?

No, the block is fixed. No it's not a homework. I am just curious and confused as well.

Then how is it accelerating?

Because of gravity. The component of gravitational Force along plane i.e gsin(theta)

It can't be both fixed and accelerating.

 

Produce a decent drawing to show what you mean.

Quote

I am sorry i apologise

Ohh i am so sorry i wrote 'block'. Sorry the wedge is fixed

Edited October 5, 20187 yr by Shahroze

10 minutes ago, Shahroze said:

Because of gravity. The component of gravitational Force along plane i.e gsin(theta)

well gsin(theta) is an acceleration so you can apply Newton's second law yes?

No i mean about the Acceleration vertically downward. Checkout the figure

3 minutes ago, Shahroze said:

Fbd of wedge,

Edited October 5, 20187 yr by Shahroze

1 hour ago, studiot said:

If there is relative motion between the block and the wedge then you must apply Newton's Laws resolved parallel and perpendicular to the interface.

It is not then an equilibrium question.

 

 

You tell me that something is moving, so why are you not applying Newton's Laws of motion?

1 minute ago, studiot said:

 

You tell me that something is moving, so why are you not applying Newton's Laws of motion?

Please check the figure i have posted i  have applied Newton's laws.

 

1 hour ago, Shahroze said:

Fbd of wedge,

I don't see them mentioned once.

 

Have you not learned how to lay out your work so that others can follow what you are doing?

 

What have i not mentioned?. I have clearly mentioned all the forces on the wedge and the components. Is it not clear to you?

No it's not clear.

And your diagram is not a free body diagram of the wedge, despite you labelling it as such.

A free body diagram does not include both forces in an N3 pair.

If the wedge is regarded as fixed then it has two N3 pairs or one N3 pair and a moment pair.