The center of mass of a 0.30 kg (non-uniform) meter stick is located at its 48-cm mark. what is the magnitude of the torque (in n⋅m) due to gravity if it is supported at the 25-cm mark? (use g = 9.79 m/s2). (never include units with the answer to any numerical question.)

Explanation:

It is given that,

Mass of the meter stick, m = 0.3 kg

Meter stick is located at its 48 cm mark, l = 48 cm

We need to find the magnitude of the torque if it is supported at the 25-cm mark, l' = 25 cm

It is given by :

So, the magnitude of the torque is 0.675 N-m. Hence, this is the required solution.

the correct choice is a .

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see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west