yes, you get a negative answer but that negative is there just to show direction. So in this case the net force in Y is directed downward.
So Fy = -3.3*10^4 (j) j = y direction
Fx = -8.81 * 10^4 (i) i = x direction (I put the negative since the forces are pulling to the left...
a) to find the magnitude you take the square root of the sum of the squared components
b) show that the angle between those vectors is 90 ( θ = cos-1 ( A dot B ) / ( |A|*|B| ) )
c) work is equal to the integral of F dot ds
d) what do you think?
Its just simple trig and breaking down the force into the components. You cannot add vectors directly, but you can add there components together.
Sum of forces in Y direction = F1sin(30) - F2sin(60) //the minus is due to opposite direction
If you translate your coordinate system, where the X axis is on the incline you will get mgsin(t) for the forces of gravity in the X direction.
[PLAIN]http://img524.imageshack.us/img524/4726/gradianthw.png
>>the force has to be greater in the "vertical" direction because the net force and acceleration goes in the direction of the greater force, right?
No, you have 1 force in the Y direction initially. Its gravity, which is pulling the plane down towards the earth. They want to know the...
what it does is that it creates an electric field in the conductor that is equivalent to the E field going through it, Those E fields cancel each other and you are left with a Net E-Field in the conductor of Zero. Than on the outside you have an E-Field that is the vector sum of the E-field...
lets go back to Trig 101, take a look
[PLAIN]http://img547.imageshack.us/img547/1426/vectors.png
sin(\theta) = Opp/Hyp
cos(\theta) = adj/Hyp
So if you look at the picture above, if you want the Y component you get
sin(\theta) = Vy/V // just replace Opp with Vy and Hyp with V
Vy =...
no problem, remember that when you have the expression for work as:
F * D
that D is the change in distance, so say you climb a mountain than you climb back down than the work overall is zero since the change in position is zero
work is
\int\vec{F}\bullet dr
to get rid of the dot product, you must take the parallel component of the distance that it is in the same direction of the force (gravity..which is down)
so you get :
-F*d*sin(\theta) = -mg*d*sin(\theta) = -(50)(9.8)*100*sin(12)
If you put a conductor in a electric field, the charge will redistribute themselves. But after this process, the E field in the conductor is not zero. There is an E-field in the conductor that is equivalent but opposite direction to the one that is passing through the conductor. That inner...
>>I know that inside a conductor the field is zero
Yes if there is no current
>>so because of the discontinuity it is \sigma/e outside
Not because of continuity? Its because the material is a conductor, therefor the charges spread themselves as far away as possible from each other, creating...