Recent content by garfiegrl

(1) ∑ T(from left end) = 0, => equilibrium condition 0.15(0.1)9.8 – 0.5F + 0.2(0.75)9.8 = 0 F = 9.8[0.015+0.15]/0.5 = 9.8(0.165/0.5) = 3.23 N (2) Interchange weights ∑ T(from left end) = 0, => equilibrium condition 0.2(0.1)9.8 – 0.43F + 0.15(0.75)9.8 = 0 F = 9.8[0.15(0.75) +...

I don't know how to do these kinds of problems; I just need an example of one, could someone just work it out with an explanation? Feel free to change the values or whatever. This particular problem isn't to turn in, I'm studying for a test on this material. A meter stick is pivoted the...

+(1.47 N)(.5m -.1 m) + -(1.96 N)(.75-.5 N) + -(50m - x)(w) = 0 ?

this? counterclockwise is +, clockwise is - right? +(1.47 N)(.5m -.1 m) + -(1.96 N)(.5-.75 N) + -(50m + x)(w) = 0 ?

+(1.47 N)(.5m -.1 m) + -(1.96 N)(.5-.75 N) + -(50m + x)(w) = 0 ?

alright. so if there's a difference of .098 N on the heavier side, and the distance on that side is .25 m, the heavier side has (0.098 N)(.25m) more torque, right? what next?

the torque on the 0-50 side is .588 N, its .490 on the 50-100 side. the 50-100 side of the ruler is greater, so the center of mass is somewhere below the 50 mark.

I'm lost...

What I did so far is this; since w1 d1 = w2 d2 I got the moments for each side on both setups, and the distance from the axis to the weight and set the equations up (1.47 N + x)(.4 m) = (1.96 + y )(.25 m) for first setup (1.96 N + x)(.33m)=(1.47 N + y)(.32m) for the second setup...

Yay! new keyboard; setup 1 w1 d1 = w2 d2 (1.47 N)(.5m -.1 m) = (1.96 N)(.5-.75 N) .588Nm = -.490Nm now what? setup 2 w1 d1 = w2 d2 (1.96 N)(.0m-.430m) = (1.47 N)(.750m -.430m) -.647 Nm = .470 Nm okay...

A meter stick is pivoted the 50-cm mark but does not balance because of nonconformities in material. 150 g and 200 g weights ar placed at the 10-cm and 75-cm marks to balance the meterstick. When the weights are interchange, the pivot point is at the 43-cm mark. find the mass of the stick...

okay, so 34 N for friction pull the box up the slope, and 98 N gravity pull down. 98 N - 34 N = 64 N needed to equalize them, and more than 64 to make it move uphill?

so the friction force is (.2)(20)(9.8)(cos 30) ? and the gravitational force is (20)(9.8)(sin 30) ? i got 34 N for friction and 98 N for gravitational. are they supposed to equal zero? or do i add them together to find the force i need to overcome? or could i just overcome the strongest?

Homework Statement Calculate the force needed to pull a mass of 20 kg at a uniform slow speed up a plane inclined at an angle of 30 with the horizontal if the coefficient of kinetic friction is 0.20. Homework Equations WN= w cos \vartheta WT= w sin \vartheta \mus= tan\vartheta The...