[SOLVED] Mechanics Review Homework - Stuck =[

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March 12th 2009, 03:59 AM
Hammeh

[SOLVED] Mechanics Review Homework - Stuck =[

A body of mass 2kg is in equilibrium under the action of three coplanar forces, P, Q and R where P=3xi-2yj N, Q=2yi-4xj N, and R=-11i+4j N.

a) Calculate the values of x and y.
The force P ceases to act. Find.
b) The resulting acceleration of the body in component form.
c) The magnitude of the acceleration, giving your answer to 1 decimal place.

i and j are unit vectors.

I am completely confused by this question, can anyone offer up a method?
Thanks!
March 12th 2009, 05:31 AM
skeeter

Quote:

Originally Posted by Hammeh

A body of mass 2kg is in equilibrium under the action of three coplanar forces, P, Q and R where P=3xi-2yj N, Q=2yi-4xj N, and R=-11i+4j N.

a) Calculate the values of x and y.
The force P ceases to act. Find.
b) The resulting acceleration of the body in component form.
c) The magnitude of the acceleration, giving your answer to 1 decimal place.

i and j are unit vectors.

I am completely confused by this question, can anyone offer up a method?
Thanks!

system is in equilibrium ... $\sum{F_i} = 0$ , $\sum{F_j} = 0$

3x + 2y - 11 = 0

-2y - 4x + 4 = 0

solve the system for x and y

P is taken away ... net force = Q + R

$a_i = \frac{F_i}{m}$ ... $a_j = \frac{F_j}{m}$

$ |a| = \sqrt{a_i^2 + a_j^2} $

do it.
March 12th 2009, 05:50 AM
e^(i*pi)

Quote:

Originally Posted by skeeter

system is in equilibrium ... $\sum{F_i} = 0$ , $\sum{F_j} = 0$

3x + 2y - 11 = 0

-2y - 4x + 4 = 0

solve the system for x and y

P is taken away ... net force = Q + R

$a_i = \frac{F_i}{m}$ ... $a_j = \frac{F_j}{m}$

$ |a| = \sqrt{a_i^2 + a_j^2} $

do it.

And once you find a you can use pythagoras to find the magnitude
March 12th 2009, 06:44 AM
skeeter

Quote:

Originally Posted by e^(i*pi)

And once you find a you can use pythagoras to find the magnitude

I thought I mentioned that ...

Quote:

$ |a| = \sqrt{a_i^2 + a_j^2} $
March 12th 2009, 07:02 AM
Hammeh

Thanks

Thanks skeeter, excellent method!
Thanks added to your profile (Rofl)

This thread can now be closed.