The Student Room Group

Mechanics inclination problem

Scroll to see replies

Reply 20
Original post by KaylaB
Well if we have the Opposite and the Adjacent of a right-angled triangle we can use Pythagoras to find the length of the Hypotenuse


Thank you so much!! 😊😊
Reply 21
Original post by Gemma_98
Thank you so much!! 😊😊


No problem :h: Happy to be of service :hat2:
Reply 22
Original post by KaylaB
I still seem to to be getting the same value for a, and thus the final answer :s-smilie:


I'll work through it in a sec and let you know.
Reply 23
Original post by Zacken
I'll work through it in a sec and let you know.


Is it not like 2:40am where you are, do you not require sleep?

Spoiler

Reply 24
Original post by KaylaB
Is it not like 2:40am where you are, do you not require sleep?

Spoiler



Ssshhh. :tongue:

Okay, so resolving down the plane: 15gsinθT=15a15g\sin \theta - T = 15a for the first particle and resolving vertically up for the second particle: T10g=10aT - 10g = 10a.

Adding these: 25a=15gsinθ10ga3.825a = 15g\sin \theta - 10g \Rightarrow a \approx -3.8 - can you take it from here?
Reply 25
Original post by Zacken
Ssshhh. :tongue:

Okay, so resolving down the plane: 15gsinθT=15a15g\sin \theta - T = 15a for the first particle and resolving vertically up for the second particle: T10g=10aT - 10g = 10a.

Adding these: 25a=15gsinθ10ga3.825a = 15g\sin \theta - 10g \Rightarrow a \approx -3.8 - can you take it from here?


Yeah I can finish it, but I have a question. I made both of those equations but I substituted in 10g-10a for T instead of adding them, how come you can just add them like that or am I just really tired?
Reply 26
Original post by KaylaB
Yeah I can finish it, but I have a question. I made both of those equations but I substituted in 10g-10a for T instead of adding them, how come you can just add them like that or am I just really tired?


If you have two equations:

a+b=ca+b = c

2a+3b=5c2a + 3b = 5c

you are allowed to substitute things in to solve. Or you can add both of them or subtract both of them or divide both of them, etc...

So if you add this is valid and you get 3a+4b=6c3a + 4b = 6c.

You can test this out: 5+3=85 + 3 =8 and 75=27-5 = 2 so 5+3+75=10=8+25 +3 + 7 -5 = 10 = 8 + 2 which is true, etc...
Reply 27
Original post by Zacken
If you have two equations:

a+b=ca+b = c

2a+3b=5c2a + 3b = 5c

you are allowed to substitute things in to solve. Or you can add both of them or subtract both of them or divide both of them, etc...

So if you add this is valid and you get 3a+4b=6c3a + 4b = 6c.

You can test this out: 5+3=85 + 3 =8 and 75=27-5 = 2 so 5+3+75=10=8+25 +3 + 7 -5 = 10 = 8 + 2 which is true, etc...


Ah okay cheers :hat2:, now rest that giant brain of yours.
Reply 28
Original post by KaylaB
Ah okay cheers :hat2:, now rest that giant brain of yours.


You two, Mrs. "am I really tired? Yes you are.". :tongue:
Reply 29
Original post by Zacken
You two, Mrs. "am I really tired? Yes you are.". :tongue:


Touché
Reply 30
Original post by KaylaB
Touché


On the other hand, I just spelt "too" as "two" so...
Reply 31
Original post by Zacken
On the other hand, I just spelt "too" as "two" so...


I was being nice and wasn't going to mention it :laugh:, you just have numbers on the brain :tongue:
Reply 32
Original post by KaylaB
I was being nice and wasn't going to mention it :laugh:, you just have numbers on the brain :tongue:


I appreciate it. Let's not make fun of Zacken the illiterate dumbo. Lol jk let's totally make fun of Zacken the illiterate dumbo, it'll be hilarious. :lol:

Quick Reply

Latest