# Question bad98

Feb 17, 2017

$M = 5 \text{ kg}$

#### Explanation:

Because the stick is suspended in the middle, the torques due to its mass are in equilibrium sum to zero in the equation, therefore, I shall not include them.

The other torques are ${\tau}_{\text{counterclockwise" and tau_}} c l o c k w i s e$"

${\tau}_{\text{counterclockwise}}$ is applied $50 \text{ cm}$ away from the center and the force is $\left(3 {\text{ kg")(-9.8" m/s}}^{2}\right)$

tau_"counterclockwise" = (50" cm")(3" kg")(-9.8" m/s"^2)

${\tau}_{\text{clockwise}}$ is applied $30 \text{ cm}$ away from center and the force is $M \left(- 9.8 {\text{ m/s}}^{2}\right)$:

tau_"clockwise" = (30" cm")M(-9.8" m/s"^2)#

Because the stick in in equilibrium, we may set ${\tau}_{\text{clockwise" = tau_"counterclockwise}}$:

$\left(30 {\text{ cm")M(-9.8" m/s"^2) = (50" cm")(3" kg")(-9.8" m/s}}^{2}\right)$

Solve for M

$M = \left(50 {\text{ cm")/(30" cm")(3" kg")(-9.8" m/s"^2)/(-9.8" m/s}}^{2}\right)$

$M = 5 \text{ kg}$