Shahroze Posted October 5, 2018 Share Posted October 5, 2018 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. Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 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. Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 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 Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 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? Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 No, the block is fixed. No it's not a homework. I am just curious and confused as well. Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 Then how is it accelerating? Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 Because of gravity. The component of gravitational Force along plane i.e gsin(theta) Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 It can't be both fixed and accelerating. Produce a decent drawing to show what you mean. Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 (edited) Quote I am sorry i apologise Ohh i am so sorry i wrote 'block'. Sorry the wedge is fixed Edited October 5, 2018 by Shahroze Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 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? Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 (edited) No i mean about the Acceleration vertically downward. Checkout the figure 3 minutes ago, Shahroze said: Fbd of wedge, Edited October 5, 2018 by Shahroze Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 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? Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 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, Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 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? Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 What have i not mentioned?. I have clearly mentioned all the forces on the wedge and the components. Is it not clear to you? Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 No it's not clear. And your diagram is not a free body diagram of the wedge, despite you labelling it as such. Link to comment Share on other sites More sharing options...
Shahroze Posted October 5, 2018 Author Share Posted October 5, 2018 Link to comment Share on other sites More sharing options...
studiot Posted October 5, 2018 Share Posted October 5, 2018 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. Link to comment Share on other sites More sharing options...
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