jnock04
Badges: 5
Rep:
?
#1
Report Thread starter 1 month ago
#1
https://isaacphysics.org/questions/s...2020_21_week36

haven't been able to do part A or C. my working for A leads me to amg/3costheta please help thanks
0
reply
user342
Badges: 12
Rep:
?
#2
Report 1 month ago
#2
Try to find costheta in terms of a and k
1
reply
jnock04
Badges: 5
Rep:
?
#3
Report Thread starter 1 month ago
#3
(Original post by user342)
Try to find costheta in terms of a and k
i think i have done this -
cos theta = (cos30)a / (1+k) a
what next though?
0
reply
user342
Badges: 12
Rep:
?
#4
Report 1 month ago
#4
(Original post by jnock04)
i think i have done this -
cos theta = (cos30)a / (1+k) a
what next though?
Hmm I'm a bit confused by your method. Maybe send a picture of your working?
0
reply
jnock04
Badges: 5
Rep:
?
#5
Report Thread starter 1 month ago
#5
(Original post by user342)
Hmm I'm a bit confused by your method. Maybe send a picture of your working?
wait i did that wrong i think it might be cos theta = a / cos(30) (1+k) a
ill get a picture of my working in a minute
0
reply
jnock04
Badges: 5
Rep:
?
#6
Report Thread starter 1 month ago
#6
here
Attached files
1
reply
user342
Badges: 12
Rep:
?
#7
Report 1 month ago
#7
That's good. You can simplify that expression a bit more, and then extend your method of balancing forces vertically to find R, the reaction force, and balance the components in horizontal direction.
0
reply
user342
Badges: 12
Rep:
?
#8
Report 1 month ago
#8
(Original post by user342)
That's good. You can simplify that expression a bit more, and then extend your method of balancing forces vertically to find R, the reaction force, and balance the components in horizontal direction.
You might need to check when resolving whether it's costheta or sintheta that's needed to find the reaction force.
0
reply
jnock04
Badges: 5
Rep:
?
#9
Report Thread starter 1 month ago
#9
(Original post by user342)
That's good. You can simplify that expression a bit more, and then extend your method of balancing forces vertically to find R, the reaction force, and balance the components in horizontal direction.
am i right to say that k^3mg = 3R*cos theta?
if i then substitute my formula for cos theta in i end up with this as the reaction force but i don't think its right:
Attached files
0
reply
user342
Badges: 12
Rep:
?
#10
Report 1 month ago
#10
(Original post by jnock04)
am i right to say that k^3mg = 3R*cos theta?
if i then substitute my formula for cos theta in i end up with this as the reaction force but i don't think its right:
Are you sure it's costheta?
0
reply
jnock04
Badges: 5
Rep:
?
#11
Report Thread starter 1 month ago
#11
(Original post by user342)
Are you sure it's costheta?
Is it sin theta then? If so how would I convert my formula I have for cos theta to sin theta to put it into the tension equation?
0
reply
user342
Badges: 12
Rep:
?
#12
Report 1 month ago
#12
(Original post by jnock04)
Is it sin theta then? If so how would I convert my formula I have for cos theta to sin theta to put it into the tension equation?
Do you know any trig laws, or pythagoras that could help you find sintheta?
0
reply
jnock04
Badges: 5
Rep:
?
#13
Report Thread starter 1 month ago
#13
(Original post by user342)
Do you know any trig laws, or pythagoras that could help you find sintheta?
Yes sorry I wasn't thinking 🤦 find sin theta using sin2theta + cos2theta = 1???
Last edited by jnock04; 1 month ago
0
reply
user342
Badges: 12
Rep:
?
#14
Report 1 month ago
#14
(Original post by jnock04)
Yes sorry I wasn't thinking 🤦 find sin theta using sin2theta + cos2theta = 1???
I personally used pythagoras to find an expression for the opposite edge, but I don't see why that wouldn't work too.
0
reply
jnock04
Badges: 5
Rep:
?
#15
Report Thread starter 1 month ago
#15
(Original post by user342)
I personally used pythagoras to find an expression for the opposite edge, but I don't see why that wouldn't work too.
tbh that sounds easier i found the opposite edge as 1/2*a + a*sqrt(2k) + ak
not sure if this is right or not because when i divide it by h (a + ak) to find sin theta and then put it in the tension formula from before i still get it wrong. sorry for this bother just really want to figure this out
0
reply
user342
Badges: 12
Rep:
?
#16
Report 1 month ago
#16
(Original post by jnock04)
tbh that sounds easier i found the opposite edge as 1/2*a + a*sqrt(2k) + ak
not sure if this is right or not because when i divide it by h (a + ak) to find sin theta and then put it in the tension formula from before i still get it wrong. sorry for this bother just really want to figure this out
I don't have my working on me atm, but I remember that the opposite edge was something like a*sqrt(k^2+2k-1/3). Do you want to check your working and post it?
0
reply
jnock04
Badges: 5
Rep:
?
#17
Report Thread starter 1 month ago
#17
(Original post by user342)
I don't have my working on me atm, but I remember that the opposite edge was something like a*sqrt(k^2+2k-1/3). Do you want to check your working and post it?
this is my working, knowing me i have probably done something stupid with the simple maths
Attached files
0
reply
user342
Badges: 12
Rep:
?
#18
Report 1 month ago
#18
2 things: check your squaring of a/cos30, and when you're sqrting (op)^2, you've got to sqrt the whole of the rhs. If you sqrt 1 side of an equation, you've got to sqrt the whole of the other side. Eg if I have 5^2=4^2+3^2 (25=16+9), 5 doesn't = 4+3, but it does = sqrt(16+9) <-- sqrt(16+9) doesn't equal sqrt(16)+sqrt(9).
Last edited by user342; 1 month ago
0
reply
jnock04
Badges: 5
Rep:
?
#19
Report Thread starter 1 month ago
#19
(Original post by user342)
2 things: check your squaring of a/cos30, and when you're sqrting (op)^2, you've got to sqrt the whole of the rhs. If you sqrt 1 side of an equation, you've got to sqrt the whole of the other side. Eg if I have 5^2=4^2+3^2 (25=16+9), 5 doesn't = 4+3, but it does = sqrt(16+9) <-- sqrt(16+9) doesn't equal sqrt(16)+sqrt(9).
oof so a/cos30 squared should be 4a^2/3 right? then i get the opposite side as sqrt(a(-1/3a +2k + ak^2)). if this is write then it is my working out after that isnt correct now ive fixed this bit?
0
reply
user342
Badges: 12
Rep:
?
#20
Report 1 month ago
#20
(Original post by jnock04)
oof so a/cos30 squared should be 4a^2/3 right? then i get the opposite side as sqrt(a(-1/3a +2k + ak^2)). if this is write then it is my working out after that isnt correct now ive fixed this bit?
Your a/cos(30) looks good to me now. Still not sure on the sqrt. Your method for finding R after seems fine.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

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

Personalise

Are you tempted to change your firm university choice on A-level results day?

Yes, I'll try and go to a uni higher up the league tables (39)
28.89%
Yes, there is a uni that I prefer and I'll fit in better (12)
8.89%
No I am happy with my choice (74)
54.81%
I'm using Clearing when I have my exam results (10)
7.41%

Watched Threads

View All