Qualification > Math

M1 HELP NEEDED, RIGHT NOW!!

<< < (7/7)

astarmathsandphysics:
d)After 7 s speed =8,4m/s
F=ma for P
uR=ma so -2/7*4g=4a so a=-2.8m/s^2
v=u+at
0=8.4-2.8a so a=3
e)F=ma
40-2/7*6g=40-2/7*6*9.8=40-16.8=23.2=6a
a=23.2/5=3.973m/s^2
v=u+at=8.4+3*3.87=20m/s

Phosu:
jun 09 5bi and ii)
i rote the reaction force as R = 1.1gCos a - PSin a
can someone plz explain how to do it?
cuz the force of P that im getting turns out to be R in the ms

halosh92:

--- Quote from: Phosu on May 21, 2010, 05:24:26 pm ---jun 09 5bi and ii)
i rote the reaction force as R = 1.1gCos a - PSin a
can someone plz explain how to do it?
cuz the force of P that im getting turns out to be R in the ms

--- End quote ---

R+ P sin 50 = 15g
R=  15g -P sin 50
P cos 50 - friction
P cos 50= 0.2 ( 15g- P sin 50)
P cos 50= 29.4- P 0.2 sin 50
P cos 50 + P 0.2sin 50 = 29.4
P( cos 50 + 0.2 sin 50) = 29.4
P= 37 N

Phosu:
im soooooooooo srry i rote the rong month, its jan 09

cooldude:

--- Quote from: halosh92 on May 21, 2010, 05:50:31 pm ---R+ P sin 50 = 15g
R=  15g -P sin 50
P cos 50 - friction
P cos 50= 0.2 ( 15g- P sin 50)
P cos 50= 29.4- P 0.2 sin 50
P cos 50 + P 0.2sin 50 = 29.4
P( cos 50 + 0.2 sin 50) = 29.4
P= 37 N

--- End quote ---

let coefficient of friction=c and alpha=x
Psinx+1.1gcosx=R
resolving horizontally (parallel to the plane)-->
Pcosx+F=1.1gsinx, F=cR, F=0.5R
Pcosx+0.5(Psinx+1.1gcosx)=1.1gsinx
Pcosx+0.5Psinx+0.55gcosx=1.1gsinx
P(cosx+0.5sinx)=1.1gsinx-0.55gcosx
P(cosx+0.5sinx)=11sinx-5.5cosx
P=2

Navigation

[0] Message Index

[*] Previous page