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# M1 Help needed! watch

1. Been doing some questions from the June 2001 paper and having some troubles.

Questions have been attached to this post.
Attached Images

2. Q2)
taking i parallel to P and j perp. to P
resolve Q: 3cos40 i + 3sin40 j
P : 5i
resultant force = (5+3cos40) i + 3sin40 j
magnitude of R = sqrt[(5+3cos40)² + (3sin40)] = 7.55 N

[email protected] = 3sin40/(5+3cos40)
@ = 14.8o
3. 2a) Qh = 3cos40
Fh = Qh + P = 3cos40+5
Fv = Qv = 3sin40
|F|2 = (3cos40+5)2 + (3sin40)2 = 56.9813...
|F| = 7.55N (3s.f.)

b) x = arctan{Fv/Fh} = 14.8 (1d.p.)
4. Ok...well the first one is as follows: (for the recors..I'm really not great at M1!)
You resolve horicontally & vertically (because the 3N force goes diagonally)
So resolve vertically: 5 (the force along the floor) + 3cos40
Resolve horizontally: 3sin40

Work out these number on calculator, and you get:
(A) 5 + 3cos40 = 7.298....
(B) 3sin40 = 1.928...

TO find the magnitude, you use pythagoras: a squared + b squared = c squared
(you do this because A is the force across, B is the force up of F - like a vector)

So... square root of (A squared + B squared) = 7.54859...

Magnitude is 7.55 (3s.f.)

b) once you know the answer to part a):

opp = 1.928...., adj = 7.29836...
then it is tan -1 (1.98../7.29..)
5. Q4)
b)
R = 30 sin30 + 3g cos30 = 40.5 N

c) f + 3g sin30 = 30 cos30
f = 11.3
f = μ R
11.3 = μ 40.5
μ = 0.28
Attached Images

6. Q6) a) F = ma
2320 - ( 800 + 240 ) = 3200 a
a = 0.4 ms-2

b) T - 240 = 1200 x 0.4
T = 720 N

c) 2320 - ( 240 + 800 + 3200g [email protected]) = 3200 a
3200 a = -288
a = -0.09 ms-2

speed decreases.
7. 6a) Using 'F=ma':
2320-800-240 = (2000+1200)a
a = 0.4ms-2

b) Resolving // to ground for van:
2320-800-T=2000(0.4)
T = 720N

c) Resolving // to slope for weight of van:
W//v = [email protected] = 2000gsin(arctan1/20) = 978.8
'' '' of car:
W//c = 1200gsin(arctan1/20) = 587.3

Using 'F=ma':
2320-800-240-W//v-W//c= (2000+1200)a
a = -0.0894ms-2(3s.f.)
Velocity decreasing, hence negative acceleration.
8. 2000gsin(arctan1/20) = 978.8
you have [email protected], why did u use sin(arcsin1/20) ? this is inacurate. sorry for saying that, although it's right. cheers.
9. Could someone please explain the 2nd quation that he posted? In the second part, are you resolving parallel to plane/perpendicular to place/horizontally/vertically? V. confused
10. all the resolving are in the same plane, so the resolving is vertica and horizontal. or parallel to force P and perp. to force P. ( which is also vertical and horizontal )
11. so why aren't they equal to one another?? WHy are they added together?
12. I mean..on your working out:
b) R = 3g sin30 + 3cos30
why is it that you resolve horizontally & vertically to get the normal reaction? Why aren't the two forces equal to eachother, because they are holding the particle there?
13. oh sorry u r talking about Q4, i thought u said Q2 not the second Question, anyway:
u resolve the forces parallel and perp to the slope. ( the red lines are the ORIGINAL forces , the BLUE [which appear blakc ] lines are the resolved forces )
anyway, see the diagram: the object is in equilibrium, parallel to the slope and perp. to the slope, so the forces perp. to the slope are equal. i.e. R = 30sin30 ( which is the resolved part of the force applied ) + 3g cos30 ( which is the resolved part of the WEIGHT)
14. Thank you so much!! I think it makes sense now..they are both perpendicular to the plane!!
15. 15 + 25.5 = 40.5N
30cos30-40.5u-3gsin30=0
11.28=40.5u
u=0.279
16. (Original post by Queen_A)
Thank you so much!! I think it makes sense now..they are both perpendicular to the plane!!
not the plane, the SLOPE, as teh plane means the plane of te system, and perp. to the plane means out of / into the page.
17. (Original post by glance)
15 + 25.5 = 40.5N
30cos30-40.5u-3gsin30=0
11.28=40.5u
u=0.279

Sorry, but what is that? what is u? which question is that from? i cant understand...
18. (Original post by yazan_l)
Sorry, but what is that? what is u? which question is that from? i cant understand...
Sorry, it was just some calculations that I'd written down. It was the second question, u was "mu" as in F=uR except I don't have the mu symbol.
19. (Original post by glance)
Sorry, it was just some calculations that I'd written down. It was the second question, u was "mu" as in F=uR except I don't have the mu symbol.
oh sorry didnt notice that i thought tht there was a part that we didnt do cheers

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