The torque is the component of the weight that is perpendicular to the bar. So when the bar hangs vertically down, parallel to the force of gravity, there is no torque. If the bar makes an angle "A" with the vertical then the component of weight perpendicular to the bar would be mgSin(A) and the torque would be mgLSin(A) , where m= 1.21 kg, and L = 1.28m, g=9.8m/ss , so all you need is the angle "A".
If the bar magnet turned then there was a torque acting upon it. Torque is defined as a turning force or moment.
vtech is having 2 cams one for low end torque then switches to the other for high end torque vtech is having 2 cams one for low end torque then switches to the other for high end torque
sway bar end linksway bar end link
Fixed bar/Torque style.
big end bearinds 1st loose torque 45nm +90 Degree then do final torque 87nm.
aku tanyo mu ni berok
the big end bearings torque down at 55 - 65 Nm
What is the big end torque setting for a fiesta st150
Torque settings for main bearings and big end bearings
65
The reaction at the fixation point(let's mark it 'A' point) will consist of reaction force and reaction torque. The net force for static item, consisting of sum of all the forces and the net torque have both to equal zero. To calculate force and torque at the fixation point, we need to know two things: mass of the bar and horizontal displacement of bar's center of mass from the fixation point. Since there are no forces acting in the horizontal plane, we can only take z(height) axis into consideration for forces. Reaction plus force of gravity have to be zero: RA + Q = 0, where Q denotes weight and is equal to Q = m * g, where m is mass of the bar. Calculated, RA = -Q, which means that reaction force is equal to weight, but directed in opposite way. Sum of all the torques in the bar has to be zero: MA + Q * l = 0, where MA is reaction torque at the fixation point, Q as above and l is distance from fixation point to bar's center of mass. Calculated: MA = -Q * l