Question #bad98

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Question #bad98

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Answer 1 · 2017-02-17T19:51:26

$M = 5 kg$ $M = 5" kg"$

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 $τ_{counterclockwise} and τ c l o c k w i s e$ $tau_"counterclockwise" and tau_"clockwise$ "

$τ_{counterclockwise}$ $tau_"counterclockwise"$ is applied $50 cm$ $50" cm"$ away from the center and the force is $(3 kg) (- 9.8 {m/s}^{2})$ $(3" kg")(-9.8" m/s"^2)$

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

$τ_{clockwise}$ $tau_"clockwise"$ is applied $30 cm$ $30" cm"$ away from center and the force is $M (- 9.8 {m/s}^{2})$ $M(-9.8" m/s"^2)$ :

$τ_{clockwise} = (30 cm) M (- 9.8 {m/s}^{2})$ $tau_"clockwise" = (30" cm")M(-9.8" m/s"^2)$

Because the stick in in equilibrium, we may set $τ_{clockwise} = τ_{counterclockwise}$ $tau_"clockwise" = tau_"counterclockwise"$ :

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

Solve for M

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

$M = 5 kg$ $M = 5" kg"$

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