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Mathematics: Post your doubts here!

Esme

Esme

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ZainH said:
Ohlord, I can't think atm
Can someone solve these 2 simultaneous equations for me.

5.5sinx = wgcosx
And
5.5cosx + wgsinx = 7.3
Click to expand...

sinx/cosx =wg/5.5
tanx * 5.5 = wg

Substitute this in the second equation
5.5cosx + sinx( tanx * 5.5) = 7.3
after simplifying the brackets, you'll get
5.5cos^2 x + 5.5 sin^2 x = 7.3 cosx
5.5( cos^2 x + sin^2 x) =7.3 cosx
5.5/7.3 = cosx
x= 41.4

now wg= tan41.4 * 5.5
 
ZainH

ZainH

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GorgeousEyes said:
Please juun012 v.42 no.5 with steps pleaase .
Click to expand...

First part is easy, since you know tension is a force use F= mass x acceleration

For S1, only weight hanging is 3kg so: F= 3x10 = 30N
For S2, both weights are hanging , so: F=5x10 = 50N

Part 2 is a bit tricky, but easy to understand.

When the masses fall , they'll have forces acting on them in different directions. For mass A it's weight will be the only thing acting downwards. Tension in S1and air resistance will be acting against it, so we can make an equation with this information.

Fnet= 30-T-1.6=3a

When mass B falls, it's weight and tension in string S1 will be acting downwards, where as air resistance will be acting upwards. It's equation is:
Fnet= 20+T-4=2a

Now solve them simultaneously,

The T's will cancel out.

30-1.6=3a
20-4=2a

44.4=5a
a= 8.88 m.s-2

Now just put your value for a back into one of the equations to get T.
20+T-4 = 2(8.88)
T = 1.76N

Hope you got it x)
 
GorgeousEyes

GorgeousEyes

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ZainH said:
First part is easy, since you know tension is a force use F= mass x acceleration

For S1, only weight hanging is 3kg so: F= 3x10 = 30N
For S2, both weights are hanging , so: F=5x10 = 50N

Part 2 is a bit tricky, but easy to understand.

When the masses fall , they'll have forces acting on them in different directions. For mass A it's weight will be the only thing acting downwards. Tension in S1and air resistance will be acting against it, so we can make an equation with this information.

Fnet= 30-T-1.6=3a

When mass B falls, it's weight and tension in string S1 will be acting downwards, where as air resistance will be acting upwards. It's equation is:
Fnet= 20+T-4=2a

Now solve them simultaneously,

The T's will cancel out.

30-1.6=3a
20-4=2a

44.4=5a
a= 8.88 m.s-2

Now just put your value for a back into one of the equations to get T.
20+T-4 = 2(8.88)
T = 1.76N

Hope you got it x)
Click to expand...

Youu are great mash'allah (y) thank youuu :)
 
ZainH

ZainH

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GorgeousEyes said:
Youu are great mash'allah (y) thank youuu :)
Click to expand...

You're welcome x)

Esme said:
sinx/cosx =wg/5.5
tanx * 5.5 = wg

Substitute this in the second equation
5.5cosx + sinx( tanx * 5.5) = 7.3
after simplifying the brackets, you'll get
5.5cos^2 x + 5.5 sin^2 x = 7.3 cosx
5.5( cos^2 x + sin^2 x) =7.3 cosx
5.5/7.3 = cosx
x= 41.4

now wg= tan41.4 * 5.5
Click to expand...

Thank you, my brain stopped e_e
 
mominzahid

mominzahid

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ZainH said:
Oh that's easy just make 2 simultaneous equations.
Use s=ut+1/2at^2

So your first equation is: 55=5u+12.5a
And second one is: 120=10u+50a

Solve them simultaneously, you should get a=0.4 and u=10 m.s-1
Click to expand...
where did the 2nd equation come from? :/
 
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