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shift3
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#1
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Is it just me, or are the statics questions in rev ex 2 of the Heinemann book exceptionally difficult?

I attempted questions: 25, 26, 27, 29, 30, 31, 34, 35 and 36.
I answered: 25, 29, 30, 31, 34 and 36... :eek:

Can anyone help me out? :|
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MdSalih
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#2
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Highlight a particular question your having a problem with.. I'll see if I can go through it

MdSalih
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shift3
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#3
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31.
I'm usually good at ladder questions, so I don't know why I messed this one up. Here's what I did:

Resolving vertically:
R1 = Mg + 10Mg = 11Mg

Resolving horizontally:
R2 = F = uR1 = (11/4)Mg (F = friction)

Moments about A:
Mg(l/2)[email protected] + [email protected] = R2(l)[email protected] = (11/4)[email protected]
(l/2) + 10x = (11/4)l [email protected]

Which doesn't give x=(6l/7)...
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ThugzMansion7
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(Original post by MdSalih)
Highlight a particular question your having a problem with.. I'll see if I can go through it

MdSalih
page 90, Q35 is hard!! any1 done this? :eek:
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shift3
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(Original post by ThugzMansion7)
page 90, Q35 is hard!! any1 done this? :eek:
Which part do you need help with?
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ThugzMansion7
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(Original post by shift3)
Which part do you need help with?
part a)
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Gaz031
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(Original post by ThugzMansion7)
page 90, Q35 is hard!! any1 done this? :eek:
(a)
Horizontally: 14cost = U
Vertically: u=14sint, s=5, a=-9.8, v=u.
v^2 = u^2 + 2as
u^2 = (14sint)^2 - 98
u^2 = 196 - 196cos^2 t - 98
u^2 - 98 = -196cos^2 t
196(cost)^2 = 98-u^2
From the horizontal equation, cost=u/14
u^2 = 98-u^2
2u^2 = 98
u^2 = 49
u = 7

(b)
cost=u/14
theta = 60 degrees

(c) u = 14sin60, a=-9.8, v=0, t= ?
v = u + at
0 = 14sin60 - 9.8t
t = 1.24s

(d) (Angle of velocity(perpendicular)with horizontal) = (90-angle of parallel with horizontal).
Hence V makes an angle 90-50=40 degrees with the horizontal.

Conservation of horizontal velocity gives:
14cos60 = Vcos40
V = 14cos60/cos40
V = 9.14m/s
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ThugzMansion7
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(Original post by Gaz031)
(a)
Horizontally: 14cost = U
Vertically: u=14sint, s=5, a=-9.8, v=u.
v^2 = u^2 + 2as
u^2 = (14sint)^2 - 98
u^2 = 196 - 196cos^2 t - 98
u^2 - 98 = -196cos^2 t
196(cost)^2 = 98-u^2
From the horizontal equation, cost=u/14
u^2 = 98-u^2
2u^2 = 98
u^2 = 49
u = 7

(b)
cost=u/14
theta = 60 degrees

(c) u = 14sin60, a=-9.8, v=0, t= ?
v = u + at
0 = 14sin60 - 9.8t
t = 1.24s

i tried that by using k.e.1 = mgh + k.e2

why didnt this work
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shift3
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(Original post by ThugzMansion7)
i tried that by using k.e.1 = mgh + k.e2

why didnt this work
At the ground:
KE=0.5mv²=98m

In the air:
KE=0.5mv²=0.5m(2u²)=mu²
PE=mgh=5mg

mu²+5mg=98m
u²=98-5g=49
u=7
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shift3
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#10
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#10
(Original post by shift3)
31.
I'm usually good at ladder questions, so I don't know why I messed this one up. Here's what I did:

Resolving vertically:
R1 = Mg + 10Mg = 11Mg

Resolving horizontally:
R2 = F = uR1 = (11/4)Mg (F = friction)

Moments about A:
Mg(l/2)[email protected] + [email protected] = R2(l)[email protected] = (11/4)[email protected]
(l/2) + 10x = (11/4)l [email protected]

Which doesn't give x=(6l/7)...
So.. Can anyone see where I went wrong?
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ThugzMansion7
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(Original post by shift3)
At the ground:
KE=0.5mv²=98m

In the air:
KE=0.5mv²=0.5m(2u²)=mu²
PE=mgh=5mg

mu²+5mg=98m
u²=98-5g=49
u=7
I don't get why u have got 2u² and not just u²
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shift3
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#12
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#12
think of it as i & j components:
v=ui+uj
speed=|v|=sqrt(u²+u²)=sqrt(2u ²)
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ThugzMansion7
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(Original post by shift3)
think of it as i & j components:
v=ui+uj
speed=|v|=sqrt(u²+u²)=sqrt(2u ²)
ah cool
thanks
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ThugzMansion7
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One very very very last question be4 i go offline, page 131 in the orange book,
Example 2-------> how comes the force of X is not opposing P? bcos Y is opposing mg
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Gaz031
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(Original post by shift3)
So.. Can anyone see where I went wrong?
It says the height above the ground, not the distance along the ladder. You need to multiply that expression by sin(theta).
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shift3
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#16
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(Original post by Gaz031)
It says the height above the ground, not the distance along the ladder. You need to multiply that expression by sin(theta).
omg!

:hmpf:
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Gaz031
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(Original post by ThugzMansion7)
One very very very last question be4 i go offline, page 131 in the orange book,
Example 2-------> how comes the force of X is not opposing P? bcos Y is opposing mg
It doesn't matter which way you draw it as it can be positive or negative (in the reverse direction).
If you look further through the working X turns out to be negative so it does in fact balance P and acts left.
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ThugzMansion7
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(Original post by Gaz031)
It doesn't matter which way you draw it as it can be positive or negative (in the reverse direction).
If you look further through the working X turns out to be negative so it does in fact balance P and acts left.
ah yes the same with collisions
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Gaz031
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(Original post by ThugzMansion7)
ah yes the same with collisions
Yes. Important to remember it's a vector rather than a scalar.
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shift3
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#20
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#20
Gaz, can you do question 35? (review ex p157)
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