How much torque must the pin exert to keep the rod in the figure from rotating?
Calculate this torque about an axis that passes through the point where the pin enters the rod and is perpendicular to the plane of the figure.
Known Variables:
L: 80 cm or .8m
M(of the rod)= 2.0 kg
m(of the weight)= 500g or .5 kg
The weight of the rod with length L acts as the center of the rod.
this means
L/2 from the pin
The torque about the pin is τ = (mg)L +(Mg)L/2
=(0.500kg x 9.8m/s2)0.8m+(2.0kg x 9.8m/s2 x 0.8m) / 2
= 3.92 + 7.84
=11.76
τ =12N*M
Leave a Reply to 日本最高級スーパーコピーブランド時計激安通販専門店,高品質時計コピー,2015最新作、国際ブランド腕時計コピー、業界唯一無二.世界一流の高品質ブランドコピー時計,当店はスーパーコピ Cancel reply