Recent content by chemboy

the reason I keep going back to this Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos0) is because that's how they answered the question in the text (solving trignometrically using cosine rule with the FBD)

so is this where I draw the FBD?? and I guess I will need to know the angles of the other vectors so q1 - q2 = 53 q3 - q1 = 90 q2 - q3 = 180 - (53 + 90) = 37 and 2F1 = k*q1*q2 / .05^2 = 2.7 X 10^-21N 3F1 = k*q3*q1 / .03^2 = 8.75 X 10^-21N 2F3 = k*q2*q3 / .04^2 =5.9 X 10^-21...

sorry I am lost here, unfortunately the book provided for this distance ed course is poorly written and the examples provided don't really help with answering of questions. what in the trig am I messing up?? seeing as this is based on the original distribution of forces, I am asuming I am...

Fnet = sqrt(2F1^2 + 3F1^2 - 2(2F1)(3F1)cos37) ?? I really need to touch up on my trig before I go back to school

sin0 = opp/hyp sin^-1(.03/.05) = 36.8 = 37 degrees

well, I believe I have found the forces acting on q1 (by q2 and q3). I guess my problem is I don't know what to do next. From the example I am given in the text they used cosine rule and a FBD to calculate the resultant of these forces. I am guessing my problem is my Trig is rusty... how...

3 electrons q1, q2, q3 setup in a right angled triangle q1 |\ | \ q3-q2 charges are q1 = +2.5 X 10^-17C q2 = +3.0 X 10^-17C q3 = +3.5 X 10^-17C distances between the charges q1-q3 = .03 m q1-q2 = .05 m setup FBD \ \ 2F1 \ / / 3F1 q1 2F1 = k*q1*q2 / .05^2 = 2.7...

find the magnitude and direction of the force acting on q1

3 electrons q1, q2, q3 setup in a right angled triangle q1 |\ | \ q3-q2 charges are q1 = +2.5 X 10^-17C q2 = +3.0 X 10^-17C q3 = +3.5 X 10^-17C distances between the charges q1-q3 = .03 m q1-q2 = .05 m setup FBD \ \ 2F1 \ / / 3F1 q1 2F1 = k*q1*q2 /...

if the following takes place simultaneously distance between two charges is doubled and one charge is doubled and the other is trippled how do I set this up mathmatically I understand that when distance doubles (r) then the effect on net charge is .25 of original F also I understand that...

The forces that are acting on the box are gravity (Fnormal) and Friction (uK) As for your first question, I am not sure what you are looking for, but Ill try: The applied force would have to overcome the objects ability to move down the ramp (Fa) and I would assume that increasing applied...

A box with a mass of 22 kg is at rest on a ramp inclined at 45 (degrees) to the horizontal. The coefficients of friction between the box and the ramp are: (mu(s): o.78 and mu(k): o.65) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular...

The question is about a skateboarder going down an incline so the book draws the triangle as a left angled triangle in the second example (if this is the proper name for this), so does this reverse the laws of cos and sin??

Im having an issue of when to use Cos and Sin. When calculating Force (Normal and Frictional) with multiple demension problems you are typically given the angle of the incline. When trying to calculate for Fgx (or basically x) I would think to use cos(theta) and when dealing with Fgy (or...

Yes it was, thanks a million :)