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Hi, could you please help me decompose these forces or guide me.
Not sure how to continue of if I'm doing it right.
Question:
https://gyazo.com/7eaac8a7447746572a108176558e338e
https://gyazo.com/e382038d31088fb188085296f1b02038
My working out:
https://gyazo.com/67a321fde626a2ef2890347be0682606
Thanks!
Not sure how to continue of if I'm doing it right.
Question:
https://gyazo.com/7eaac8a7447746572a108176558e338e
https://gyazo.com/e382038d31088fb188085296f1b02038
My working out:
https://gyazo.com/67a321fde626a2ef2890347be0682606
Thanks!
Last edited by Jian17; 1 week ago
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#2
(Original post by Jian17)
Hi, could you please help me decompose these forces or guide me.
Not sure how to continue of if I'm doing it right.
Question:
https://gyazo.com/7eaac8a7447746572a108176558e338e
My working out:
https://gyazo.com/67a321fde626a2ef2890347be0682606
Thanks!
Hi, could you please help me decompose these forces or guide me.
Not sure how to continue of if I'm doing it right.
Question:
https://gyazo.com/7eaac8a7447746572a108176558e338e
My working out:
https://gyazo.com/67a321fde626a2ef2890347be0682606
Thanks!
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(Original post by Physics Enemy)
Looks good, I think there's also a reaction force at B pointing diagonally left upward. I assume the pieces have negligible mass.
Looks good, I think there's also a reaction force at B pointing diagonally left upward. I assume the pieces have negligible mass.
Why would the reaction force at B be diagonally left upward if you don't mind explaining to me?
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#4
(Original post by Jian17)
I see, yes they don't give me information about the mass so it's negligible.
Why would the reaction force at B be diagonally left upward if you don't mind explaining to me?
I see, yes they don't give me information about the mass so it's negligible.
Why would the reaction force at B be diagonally left upward if you don't mind explaining to me?
Edited: Yes, along the line CB is 50° to vert (oops).
Last edited by Physics Enemy; 1 week ago
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(Original post by Physics Enemy)
Well look at incline AB, a reaction force perpendicular to it at B, points diagonally left upward (30° to vert).
Well look at incline AB, a reaction force perpendicular to it at B, points diagonally left upward (30° to vert).
Would you mind helping me with questions b and c? For b) what I did was try to do moments about
B but I have to unknowns which would be the force in pin b and c.
For a) to solve it I did moments about the left wheels for the right wheel and the force of 6000N and then with the answer substitute and find the other force exerted on the left wheel. It's correct but I'm not sure why I don't need to consider the forces on the pins, is it because I only need to consider the external forces?
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#6
(Original post by Jian17)
https://gyazo.com/e382038d31088fb188085296f1b02038
Would you mind helping me with questions b and c? For b) what I did was try to do moments about
B but I have to unknowns which would be the force in pin b and c.
https://gyazo.com/e382038d31088fb188085296f1b02038
Would you mind helping me with questions b and c? For b) what I did was try to do moments about
B but I have to unknowns which would be the force in pin b and c.
b) Taking moments about B won't work since the force of the piston is along CB, i.e it's line of action is through B, and hence the moment of that force is 0.
Try taking moments about A for the inclined rod AB.... The only forces you need to take account of are the 6000N force and the force due to the piston. That will give you the force acting on the piston.
I presume the
is some form of notation relating to the area of the head of the piston - not come across it.Use that to then determine the pressure.
c)
For the rod AB, you can either resolve vertically and horizontally, or use a triangle of forces. Which ever method you're familiar with.
For a) to solve it I did moments about the left wheels for the right wheel and the force of 6000N and then with the answer substitute and find the other force exerted on the left wheel. It's correct but I'm not sure why I don't need to consider the forces on the pins, is it because I only need to consider the external forces?
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(Original post by ghostwalker)
If treat the whole apparatus as one thing, you only need to consider external forces. Internal forces will cancel out - e.g. The force of AB on the base at A is equal and opposite (Newton's 3rd law) to the force of the base on AB at A.
If treat the whole apparatus as one thing, you only need to consider external forces. Internal forces will cancel out - e.g. The force of AB on the base at A is equal and opposite (Newton's 3rd law) to the force of the base on AB at A.
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#8
(Original post by Jian17)
I managed to do part b thanks a lot. For part c would the force on pin A act along AB? Not sure why correct answer is 5787N at 12.6degrees to the horizontal, shouldn't the angle be 30degrees?
I managed to do part b thanks a lot. For part c would the force on pin A act along AB? Not sure why correct answer is 5787N at 12.6degrees to the horizontal, shouldn't the angle be 30degrees?
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