Recent content by Natbird

k=F/dx where dx=distance package is away from center at the gravitational Force F found earlier angular frequency of package=sqrt(k/mass of package) 2pi/angular frequency of package=T=total time is this correct

Part1-A tunnel is bored through the center of a planet, as shown in the figure. (This drawing is NOT to scale and the size of the tunnel is greatly exaggerated.) Assume that the planet is a homogenous sphere with a total mass M = 3.5x 1024 kg and a radius R = 6600 km. A package of mass m = 7.5...

I think I'm going to change my axis of rotation to make it a little simplier. the orgin for the axis will be the spot where the plank touches the ground. This should leave the only unknown F since the distance from N is 0 setting N to zero since Torque=FxR next the force due to the weight...

if I take the orgin as the spot where the plank touches the ground would that be easier sum of torques=Fsin59*+Fcos59*(h)/sin59*-mgcos59*(1.7)=0

I think the sum of the torques is actually equal to the equation Fsin59*(h)-N(h)/sin59*[+]mgcos59*(1.7)=0 since if I take the axis from the point I did the plank would rotate counterclockwise

taking the axis at the ground below the roller sum of x direction=Fsin59*-friction=0 sum of y direction=Fcos59*+N-mg=0 sum of torque=Fsin59*(h)-N(h)/sin59*-mgcos59*(1.7)=0 solve for F by setting the N= to mg-Fcos59* are those formulas correct??

I'm confused how is a cylinder roller different in this case then? it it's Normal force mgcos59*?

do I take into consideration a horizonal force from the roller and a vertical force using sums of those vectors as the final force by the roller on the plank? http://smg.photobucket.com/albums/v231/er1smesp00n/?action=view&current=angles.jpg am I even close?

I really don't see it. Here's my diagram- the blue arrows show my constant changing of the Force from the roller vector http://smg.photobucket.com/albums/v231/er1smesp00n/?action=view&current=physicsroller.jpg

I don't know the direction of the force roller I thought it was only in the x direction because it was frictionless then I thought perhaps there's an upwards force also because the plank is not merely resting against it but resting on it is it something to do with the fact it's a roller? is...

so i assum then that the roller has a force F(roller) along the y-axis in the same direction as the Normal force Sum of Forces in y direction=Normal+Force or Roller along Y - mg=0 Sum of Forces in x direction should remain the same as before Froller=friction Sum of Torque= Force of roller...

The question: A plank, of length L = 3.4 m and mass M = 22.0 kg, rests on the ground and on a frictionless roller at the top of a wall of height h = 1.60 m (see the figure). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ >= 59° but...