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    Calculating Fnet

    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...
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    Conservative force on XY-Plane

    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?
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    Is this problem do able?

    you only need conservation of momentum
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    Calculating Fnet

    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
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    Force exerted by one object on another?

    no I think he means contact forces. You sum the forces in each direction, then solving for the force that you wish to find. Draw a Force Diagram
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    Normal force

    your Fn is correct
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    Finding force w/o mass

    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]
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    Oh man, I don't understand anything about forces Dx

    >>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...
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    Quick Electric Field Question!

    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...
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    Vectors- 1 Questions

    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 =...
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    Checking to see if I'm right or not.

    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
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    Checking to see if I'm right or not.

    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)
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    Vectors- 1 Questions

    were not a free homework service... its really not a hard question, do some research
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    Quick Electric Field Question!

    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...
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    Quick Electric Field Question!

    >>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...
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    Can anyone help me out with this tension problem please?

    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?
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    Static Friction / Need Help Solving for Theta

    Yes, you cant simply say Fs = (uk) * N. Which i see you did
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    Static Friction / Need Help Solving for Theta

    static friction is not like kinetic friction for kinetic you can say: f_{k} = \muk * N for static it is : f_{s} <= \mus * N
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    Apparent Weight and Newtons Second law Problems

    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
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    Two stacked objects sliding down inclined plane.

    >>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
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    Gravitational Potential energy

    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
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    Physics-gravitational forces

    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.
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    Calculating the electric field due to a wire of finite length

    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
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    Overcoming Friction Sliding Box.

    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
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    Force and Motionfinding tension?

    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
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    Force and Motionfinding tension?

    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
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    Need help with a torque problem.

    torque = R X F = I*w since force is acting perpendicular to the branch it can be rewritten as R*F = I*w
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    Overcoming Friction Sliding Box.

    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...
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    Trouble with Projectile Motion

    this is for 1D motion and you travel at a constant velocity, so keep that in mind
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    Friction and force at an angle

    >>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).
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