Recent content by tongpu

just do sum of the moments or torques at the bottom of the pole to get one equation one unknown consisting of Tension...

you actually want the partial derivative of r

\Sigma\tau=Wcos\theta+wcos\theta-Fsin\theta=0 using right hand rule should all be positive... or all "into the page" your angles seems to be correct. \Sigma\tau=ncos\theta-wsin\theta-fcos\theta=0 signs are correct, fcos(theta) for friction force, sin theta... and w cos theta for beam...

well for part (b)F=\sqrt{W^2+\mu_s^2(w+W)^2} if you do sum of forces in y direction n-w-W=0 n=w+W and Fs=\mu_s*n then sum of forces in x-direction you get T=\mu_s*n so resultant force at point P = magnitude of the components W and Fs. which is the same as the book. for resultant force at the...

To clear confusion, the reaction force exerted by the floor is straight up even if it is angled. The x-component of the floor on the beam is friction force given by \mu_s*normal force. The horizontal rope can be cut and replaced with a force in tension.

well first equation is F = K(q1)(q2)/r^2 and second equation is q1q2=13.4 μC 2 equations 2 unknowns.

What you want to do is express everything in impedance (Z). Since Vs is in cosine form ... phasor of it would be 25{-30 or 25cos30 - jsin30 (if { means phasor) Then you want Z of c and r in the circuit.. given by the equations.. Zr = R and Zc= -1/(jwc) Since impedances are expression...

well coulomb force can be expressed as F=qE , draw a force diagram like in part a and sum of forces equal zero...

well if you do not know mesh currents, then use Kirchoff voltage law to make 3 equation with 3 unknowns

a) should be correct b) what you want is the magnitude of the uniform electric field, not the field produced by the point charge, might want to use Newton third law to relate columb force to tension in part a c) if you remove the string what are the forces acting on the sphere?

Uh are you sure you are not suppose to find the electric field at a given point? Also maybe placing the square wire in the x-y plane will make it simpler. Regarding needing three variables k is constant, r varies (a function of a variable ex. X), and you take a small element of charge dQ (which...

I think you should break the magnetic field into components becasue 72 degree is not the angle between ILXB.

For this case dr it is just the integration operator. And since you want from V(0m) - V(.48m) = integral of E dot dr, your limits would be 0 -> .48m (you might have to split it into two integrals, since you cannot have 1/0)

Your E is wrong missing R squared. "But what do you multiply that by??" how do you simplify a dot product? (hint: you can choose any path for dr, and you know how the vector field E looks like)

Use gauss law to find E, the "r/2 and infinity" are just different limits of integration.