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All Mechanics DOUBTS HERE!!!

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hesho21:
Guys plz question no.3 in the attached paper here . It made me madd!! Thanx in advance

nid404:
Here you go....

Since the system is in equilibrium, the sum of horizontal components should be equal to zero.
That is

W2sin60o- W1sin40= 0   Call this eq 1

The sum of vertical components should also be zero         
W1cos40+ W2cos60 - 5N= 0                            Call this eq 2

Now you rearrange eqn 1 to get either W1 in terms of W2 or the other way round. I'll go with the first

Eq 1 W2sin60o= W1sin40
W1=W2sin60o/ sin40

Substitute W1 in eqn 2

W2sin60o/ sin40 (cos 40)  + W2cos60= 5
 1.53 W2= 5
  W2=3.26N

Now substitute the value of W2 in eq 1

W1sin40= 3.26 sin 60
W1=4.4N

immortal:
Ur method is confusing,what happens if dey r not in equilibrium ???
By the way is it ok 2 solve this using da data 2 form a triangle.
V wud get   5/sin80=W1 /sin60=W2/sin40.

nid404:
You can use the triangle method too...

hesho21:
Thanks sooo much that really helped  :D and ure explanation was great, good luck :D

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