# vectors and mass

• Jul 15th 2006, 09:03 AM
dopi
vectors and mass
A particle of mass m kg is acted on by two forces F1 and F2 with magnitudes 3root5 newtons and root5 newtons ad drections parallel to the vectors i+2j and i-2j respectively

The particle is initially at a position given by vector 2i+j

iv calculated the cartesian components of F1 and F2
F1 = 3i+6j
F2 = i 2j

iv also calculated the cartesian component of the total force F1+F2 = (3i+6j)+(i-2j) = 4i +4j

however the part im struggling on is that now the particle is of mass 1kg and is initially at rest, i have to use newtons 2nd law to write down an equation of motion and find the position of the particle after 1second.

so far iv tried:
F=ma
F1+F2= 1a
therefore a = 4i+4j but then i dont know how to find the new posistion ...can anyonehelp thankz
• Jul 15th 2006, 09:08 AM
malaygoel
Quote:

Originally Posted by dopi
A particle of mass m kg is acted on by two forces F1 and F2 with magnitudes 3root5 newtons and root5 newtons ad drections parallel to the vectors i+2j and i-2j respectively

The particle is initially at a position given by vector 2i+j

iv calculated the cartesian components of F1 and F2
F1 = 3i+6j
F2 = i 2j

iv also calculated the cartesian component of the total force F1+F2 = (3i+6j)+(i-2j) = 4i +4j

however the part im struggling on is that now the particle is of mass 1kg and is initially at rest, i have to use newtons 2nd law to write down an equation of motion and find the position of the particle after 1second.

so far iv tried:
F=ma
F1+F2= 1a
therefore a = 4i+4j but then i dont know how to find the new posistion ...can anyonehelp thankz

You have got the acceleration right.
You can use the equation$\displaystyle x_f=x_i + v_it + \frac{1}{2}at^2, v_i=0, t=1$
Substituting the value, you get
$\displaystyle x_f=4i + 3j$

Keep Smiling
Malay
• Jul 15th 2006, 09:13 AM
dopi
.thankz
.........................thankz