Question #bad98
↳Redirected from
"How do you name branched alkanes?"
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_"counterclockwise" and tau_"clockwise#"
#tau_"counterclockwise"# is applied #50" cm"# away from the center and the force is #(3" kg")(-9.8" m/s"^2)#
#tau_"counterclockwise" = (50" cm")(3" kg")(-9.8" m/s"^2)#
#tau_"clockwise"# is applied #30" cm"# away from center and the force is #M(-9.8" m/s"^2)#:
#tau_"clockwise" = (30" cm")M(-9.8" m/s"^2)#
Because the stick in in equilibrium, we may set #tau_"clockwise" = tau_"counterclockwise"#:
#(30" cm")M(-9.8" m/s"^2) = (50" cm")(3" kg")(-9.8" m/s"^2) #
Solve for M
#M = (50" cm")/(30" cm")(3" kg")(-9.8" m/s"^2)/(-9.8" m/s"^2)#
#M = 5" kg"#
