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Mechanics Help!
2 questions are driving me crazyyyy
any help appreciated
1)A light inextensible string- 50 cm has its upper point fixed at A and carries a particle- 8kg at lower end.
A horizontal force P applied to the particle keeps it in equilibrium 30 cm from the vertical through A.
By resolving vertically and horizontally find the magnitude of P and the tension in the string.
any help appreciated
1)A light inextensible string- 50 cm has its upper point fixed at A and carries a particle- 8kg at lower end.
A horizontal force P applied to the particle keeps it in equilibrium 30 cm from the vertical through A.
By resolving vertically and horizontally find the magnitude of P and the tension in the string.
Scroll to see replies
MIZZ
2 questions are driving me crazyyyy
any help appreciated
1)A light inextensible string- 50 cm has its upper point fixed at A and carries a particle- 8kg at lower end.
A horizontal force P applied to the particle keeps it in equilibrium 30 cm from the vertical through A.
By resolving vertically and horizontally find the magnitude of P and the tension in the string.
any help appreciated
1)A light inextensible string- 50 cm has its upper point fixed at A and carries a particle- 8kg at lower end.
A horizontal force P applied to the particle keeps it in equilibrium 30 cm from the vertical through A.
By resolving vertically and horizontally find the magnitude of P and the tension in the string.
Come on. This is pretty easy. Resolve horizontally and vertically.
MIZZ
anybody?
is it just the diagram you are having problems with?
MIZZ
it coudl be the diagram which is incorrect
heres what i;v drawn
point A
vertical line downwards - 8kg and 50 cm
horizontal line P- unknown
and a diaganol between lines P and A - 30cm
heres what i;v drawn
point A
vertical line downwards - 8kg and 50 cm
horizontal line P- unknown
and a diaganol between lines P and A - 30cm
yeah i'm sure that's wrong....the diagram is....from A the string diagonally down 50cm, also from A a vertical line straight down which goes down to the horizontal level that the mass at the end of the string is on...and thus the horizontal line from the mass to the vertical line is 30cm...this makes a 3,4,5 triangle.
king of swords
yeah i'm sure that's wrong....the diagram is....from A the string diagonally down 50cm, also from A a vertical line straight down which goes down to the horizontal level that the mass at the end of the string is on...and thus the horizontal line from the mass to the vertical line is 30cm...this makes a 3,4,5 triangle.
Then don't forget to apply all the forces to the diagram.
This is a very easy question. I agree with the last notion that in a triangle, the diagonal line is 50cm, the horizontal is 30 and the vertical is 40cm.
Hence, just find the vertical component of the tension T in the string, then find the horizontal component to be able to give you the force P, combine both the horiz and vert component and you get the tension in the string...
Hence, just find the vertical component of the tension T in the string, then find the horizontal component to be able to give you the force P, combine both the horiz and vert component and you get the tension in the string...
MIZZ
2 questions are driving me crazyyyy
any help appreciated
1)A light inextensible string- 50 cm has its upper point fixed at A and carries a particle- 8kg at lower end.
A horizontal force P applied to the particle keeps it in equilibrium 30 cm from the vertical through A.
By resolving vertically and horizontally find the magnitude of P and the tension in the string.
any help appreciated
1)A light inextensible string- 50 cm has its upper point fixed at A and carries a particle- 8kg at lower end.
A horizontal force P applied to the particle keeps it in equilibrium 30 cm from the vertical through A.
By resolving vertically and horizontally find the magnitude of P and the tension in the string.
MIZZ
hmm
so that would mean resolving vertically gives 40
and horizontally gives 30
so that would mean resolving vertically gives 40
and horizontally gives 30
no...30 and 40 are just lengths...you are after the forces...the whole point of working out the lengths is so you know what angle θ (theta) is.
Resolving the forces should give you:
P = Tcosθ
8*9.8 = Tsinθ
where sinθ = 4/5 and cosθ = 3/5
MIZZ
king of swords
still there?
still there?
yup
MIZZ
sorry to keep bothering you about the question
but were the figuresabove for theta and tension correct?
but were the figuresabove for theta and tension correct?
yes.
king of swords
yup
Can u help me out with a question Im having problems with please.
"A crate of mass 250kg rests in equilibrium on a slope inclined at 20 degrees to the horizontal. Find the frictional force acting on the crate."
When I resolve perpendicul to the plane downwards, I get 250gcos20. When I resolve downwards along the plane I get 250gsin20.
But how do I get the frictional forces?
2776
Can u help me out with a question Im having problems with please.
"A crate of mass 250kg rests in equilibrium on a slope inclined at 20 degrees to the horizontal. Find the frictional force acting on the crate."
When I resolve perpendicul to the plane downwards, I get 250gcos20. When I resolve downwards along the plane I get 250gsin20.
But how do I get the frictional forces?
"A crate of mass 250kg rests in equilibrium on a slope inclined at 20 degrees to the horizontal. Find the frictional force acting on the crate."
When I resolve perpendicul to the plane downwards, I get 250gcos20. When I resolve downwards along the plane I get 250gsin20.
But how do I get the frictional forces?
You have three forces at work...weight, reaction force and friction.
Equilibrium of forces // to the plane give you:
0 = 250gsin20 - Friction
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