# M2 staticsWatch

Announcements
This discussion is closed.
#1
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...

Can anyone help me out? :|
0
14 years ago
#2
Highlight a particular question your having a problem with.. I'll see if I can go through it

MdSalih
0
#3
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)

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)...
0
14 years ago
#4
(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?
0
#5
(Original post by ThugzMansion7)
page 90, Q35 is hard!! any1 done this?
Which part do you need help with?
0
14 years ago
#6
(Original post by shift3)
Which part do you need help with?
part a)
0
14 years ago
#7
(Original post by ThugzMansion7)
page 90, Q35 is hard!! any1 done this?
(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
0
14 years ago
#8
(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
0
#9
(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
0
#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)

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?
0
14 years ago
#11
(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
0
#12
think of it as i & j components:
v=ui+uj
speed=|v|=sqrt(u²+u²)=sqrt(2u ²)
0
14 years ago
#13
(Original post by shift3)
think of it as i & j components:
v=ui+uj
speed=|v|=sqrt(uÂ²+uÂ²)=sqrt(2uÂ ²)
ah cool
thanks
0
14 years ago
#14
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
0
14 years ago
#15
(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).
0
#16
(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!

0
14 years ago
#17
(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.
0
14 years ago
#18
(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
0
14 years ago
#19
(Original post by ThugzMansion7)
ah yes the same with collisions
Yes. Important to remember it's a vector rather than a scalar.
0
#20
Gaz, can you do question 35? (review ex p157)
0
X
new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Cardiff Metropolitan University
Sat, 19 Oct '19
• Coventry University
Sat, 19 Oct '19
• University of Birmingham
Sat, 19 Oct '19

### Poll

Join the discussion

#### Why wouldn't you turn to teachers if you were being bullied?

They might tell my parents (11)
6.01%
They might tell the bully (18)
9.84%
I don't think they'd understand (32)
17.49%
It might lead to more bullying (70)
38.25%
There's nothing they could do (52)
28.42%