physics help please!!!

[ullahabd]

please guys as i suck at physics cud u help me out in oct/nov 2010 p33 question 2b :fool:

 

[allysaleemally]

remember whenever they tell u to find the force (weight) of something u have to solve it by calculating moments about a same point.

so here the moment produced by the 160 N rod is equal to the moment produced by the weight of the man ABOUT THE PLANK P.

therefore taking the distances

0.2 multiplied by weight W = 160 multiplied by (0.5 + 0.25):

by solving this u will get the answer

 

[ullahabd]

remember whenever they tell u to find the force (weight) of something u have to solve it by calculating moments about a same point.

so here the moment produced by the 160 N rod is equal to the moment produced by the weight of the man ABOUT THE PLANK P.

therefore taking the distances

0.2 multiplied by weight W = 160 multiplied by (0.5 + 0.25):

by solving this u will get the answer

thanks for da answer bro but y does the marking scheme say the final ans 730/734

 

[ullahabd]

dis question is literally killing me!!! :%) :crazy:

 

[haochen]

which one ???
post the link of the paper
ill try to help u

 

[ullahabd]

question 2b
http://www.xtremepapers.me/CIE/Cambridg ... _qp_33.pdf

 

[unplugged]

question 2b
http://www.xtremepapers.me/CIE/Cambridg ... _qp_33.pdf

First you need to sort out the distances. The way that it's tilting you realize that the pivot is the left side of the P support.

Distance from center of floorboard to the center of the P support is (as in fig 2.1) = 0.75m
Distance from end of floorboard to the center of the P support is = 0.2m
Thickness of pivot is 0.06m

so distance from center of floorboard to pivot is 0.75 + 0.03(half of the thickness) = 0.78 m
Distance of edge of floorboard to pivot is = 0.2 - 0.03 = 0.17

Since they're balanced , the equation is: 160 x 0.78 = W x 0.17 , so W = 734N.
And the total force exerted from the pivot on the table now is everything above it which is 160(floorboard) + 734(The guy) = 894N.
And of course the rest of the forces are 0s as they're not even in contact with the floorboard.

 

[ullahabd]

thanks man :Bravo: :Yahoo!:
dis question has bin bugging me from a long time!

 

[rockincrew]

can any1 tell me how to solve q1(iii) in the above
cant get d idea

 

[xIshtar]

can any1 tell me how to solve q1(iii) in the above
cant get d idea

You see that it is a trapezium, and you know the distance is 100m..

So 8x3 = 24
/2 = 12

100 - 12 = 88

For the area of the trapezium = 1/2 (sum of parallel sides) x height

So, h = 11.2, sum of sides = 8 + x
4 + x/2* 11.2 = 44.8 + 5.6x

=> 5.6x = 88 - 44.8 = 43.2

=> x = 7.71m/s (3sf)

 

[rockincrew]

thnx

 

[haochen]

i wasnt here when u asked thoose question

 

[xIshtar]

thnx

No problems...

i wasnt here when u asked thoose question

Good luck in Chemistry today haochen

 

[haochen]

thx
gud luck to u too

 

[xIshtar]

thx
gud luck to u too

 

[rockincrew]

guyz didnt know u have chemistry
good luck to all

 

[xIshtar]

guyz didnt know u have chemistry
good luck to all

Thanks, you too

 

[haochen]

thxs