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 [Broken]
>>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 [Broken]
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...
Sum the forces in X and Y direction and solve for Ft. Its easier if you project the X axis onto the inclined plane.
Lets just Isolate one of the blocks:
[PLAIN]http://img137.imageshack.us/img137/7338/blockeah.png [Broken]
Can you sum the forces on this block?
what about the other block?
you need to get into the habit of creating Free body diagrams to show all the forces on your system, these will help you understand the physics of more complex problems in the future.
1)
Force in Y = -mg - ma = - Weight
2) again sum the forces in the x direction
>>Well, my understanding is that if there was no friction between the boy and the top of the scale
no, there is no frictional force between the scale and the ice hill
>>this must mean he's not moving
yes, relative the the scale he is not moving
this is the expression for potential energy close to earth : mgr
m = mass
g = gravity
r = height (alititude)
you are simply replacing mg with a more general expression for the forces of gravity\frac{Gm1m2}{r^{2}}*r
this expression gives you potential energy
were talking magnitudes here, force is not a vector in your case. so your force vector is
F = (55*9.8)j
|F| =\sqrt{539^{2}} = 539
that might seem irrelevant, but when your forces have two components thats the process you need to do to find the magnitude.
dL should be dy
you need to break the E field into components
the y component of the E field is (k*dq)/(x^2+y^2) * sin(t)
the x component of the E filed is (k*dq)/(x^2+y^2) * cos(t)
once you get there, use a U substitution and the integral should be trivial
take a look
[PLAIN]http://img718.imageshack.us/img718/8605/blockvg.png [Broken]
Fy = Fn - Fgcos(t) = 0
Fx = Fgsin(t) - Fk = ma
Fk = (uk)Fn = (uk)Fgcos(t)
Fx = Fgsin(t) - (uk)Fgcos(t) = ma
T does not equal time, it is tension. Maybe i should of put Ft.
So you solved for T right, now you your acceleration and you know gravity. Plug in values
you need to sum the forces; the sum of those forces must equal the net Force
Sum of the Forces in Y = T - mg = ma; where a is the acceleration of the whole system
when you do these problems, put your x axis flush with the ramp, that way the sum of the forces in the Y direction equals zero. It simplifies the problem.
[PLAIN]http://img710.imageshack.us/img710/6459/blockp.png [Broken]
If you do that than the forces in the Y = 0
Forces in Y = N - mg = 0...
>>T-F=m1a
where is your friction, and F has two components ... you need to break them up.
It would look something like this in the X
Forces in X = Fpcos(t) - T- Ff = (m1)a;
you can find friction by finding the normal force (sum the forces in the Y).