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Mechanics inclination problem

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1. Qu 1. Find the time interval between a particle reaching the bottom of a smooth slope of length 5m and inclination 1 in 98, and another particle reaching the bottom of s smooth slope of length 6m and incline 1 in 70. Both particles are released from rest at the top of their respective slopes at the same time.

Qu 2. A mass of 15kg lies on a smooth plane of inclination 1 in 49. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope, over a smooth pulley fixed at the top of the plane and has freely suspended a mass of 10kg at its other end.
If the system is releases from rest, find the acceleration of the masses and the distance each travels in the first 2s. Assume that nothing impedes the motion of either mass.

Any attempts are welcome!!
2. (Original post by Gemma_98)
Qu 1. Find the time interval between a particle reaching the bottom of a smooth slope of length 5m and inclination 1 in 98, and another particle reaching the bottom of s smooth slope of length 6m and incline 1 in 70. Both particles are released from rest at the top of their respective slopes at the same time.

Qu 2. A mass of 15kg lies on a smooth plane of inclination 1 in 49. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope, over a smooth pulley fixed at the top of the plane and has freely suspended a mass of 10kg at its other end.
If the system is releases from rest, find the acceleration of the masses and the distance each travels in the first 2s. Assume that nothing impedes the motion of either mass.

Any attempts are welcome!!
q1 well start with a diagram with 98 as length on bottom and 1 as height and fill in what you know... wait i'm confused :/

i thought 1 in 98 meant that 1 is the height and 98 is the length of the flat bit of the slope touching the ground.... but how does that work if the slope length is 5m??? :/
3. (Original post by thefatone)
q1 well start with a diagram with 98 as length on bottom and 1 as height and fill in what you know... wait i'm confused :/

i thought 1 in 98 meant that 1 is the height and 98 is the length of the flat bit of the slope touching the ground.... but how does that work if the slope length is 5m??? :/
It just means the angle is - you can help out with the rest? I'm off to bed now.
4. (Original post by Zacken)
It just means the angle is - you can help out with the rest? I'm off to bed now.
oh... i haven't done M2 or C3 or 4 yet :/
5. (Original post by thefatone)
q1 well start with a diagram with 98 as length on bottom and 1 as height and fill in what you know... wait i'm confused :/

i thought 1 in 98 meant that 1 is the height and 98 is the length of the flat bit of the slope touching the ground.... but how does that work if the slope length is 5m??? :/

I think the 1 in 98 is suppose to mean for every 1m it climbs its 98 along the bottom which allows you to figures out the angle to be
Sin theta = o/h = 1/98
6. once again my good intentions are useless ;( ;(
7. (Original post by Gemma_98)
I think the 1 in 98 is suppose to mean for every 1m it climbs its 98 along the bottom which allows you to figures out the angle to be
Sin theta = o/h = 1/98
The answer is a = 3.8ms2 and distance = 7.6m
8. (Original post by thefatone)
oh... i haven't done M2 or C3 or 4 yet :/
Somebody else will take care of it or I'll do it in a bit if nobody has yet. OP may wish to express his/her own thoughts in the mean time.
9. (Original post by Gemma_98)
I think the 1 in 98 is suppose to mean for every 1m it climbs its 98 along the bottom which allows you to figures out the angle to be
Sin theta = o/h = 1/98
Nah, tan theta = opp/adj = 1/98.

To start off: resolve down the plane, find accleeration (mg sin theta) then use SUVAT.
10. (Original post by Zacken)
Somebody else will take care of it or I'll do it in a bit if nobody has yet. OP may wish to express his/her own thoughts in the mean time.
i wish i knew more... ok then
11. The answer to qu1 is 0.74 secs but i dunno how this is
12. (Original post by Gemma_98)
Qu 1. Find the time interval between a particle reaching the bottom of a smooth slope of length 5m and inclination 1 in 98, and another particle reaching the bottom of s smooth slope of length 6m and incline 1 in 70. Both particles are released from rest at the top of their respective slopes at the same time.

Qu 2. A mass of 15kg lies on a smooth plane of inclination 1 in 49. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope, over a smooth pulley fixed at the top of the plane and has freely suspended a mass of 10kg at its other end.
If the system is releases from rest, find the acceleration of the masses and the distance each travels in the first 2s. Assume that nothing impedes the motion of either mass.

Any attempts are welcome!!
(Original post by Zacken)
Nah, tan theta = opp/adj = 1/98.

To start off: resolve down the plane, find accleeration (mg sin theta) then use SUVAT.
Exactly what Zacken said,
Here's what I got for q1, if you don't understand any of the steps feel free to ask
Spoiler:
Show

Edit: Now that you've said what the answer should be, and it matches mine, make sure you know what's going on at every stage
13. (Original post by KaylaB)
Exactly what Zacken said,
Here's what I got for q1, if you don't understand any of the steps feel free to ask
Spoiler:
Show
14. (Original post by Student403)
You can keep it! It's a mess!
15. (Original post by KaylaB)
You can keep it! It's a mess!
You only realise what you had once it's gone :3
16. (Original post by Gemma_98)
The answer to qu1 is 0.74 secs but i dunno how this is
This one I'm not as confident with the answer, but here's my attempt anywho
Spoiler:
Show
17. (Original post by KaylaB)
This one I'm not as confident with the answer, but here's my attempt anywho
You've forgotten the tension in the string. Sorry!
18. (Original post by Zacken)
You've forgotten the tension in the string. Sorry!
I still seem to to be getting the same value for a, and thus the final answer
19. (Original post by KaylaB)
Exactly what Zacken said,
Here's what I got for q1, if you don't understand any of the steps feel free to ask
Spoiler:
Show

Edit: Now that you've said what the answer should be, and it matches mine, make sure you know what's going on at every stage

Thank you!
I get the resolving and SUVAT but how did tan 1/98 become the sqr root thing?
20. (Original post by Gemma_98)
Thank you!
I get the resolving and SUVAT but how did tan 1/98 become the sqr root thing?
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

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