**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/s^{2})0.8m+(2.0kg x 9.8m/s^{2 }x 0.8m) / 2

= 3.92 + 7.84

=11.76

**τ **=12N*M