A 0.102 kg meterstick is supported at its 37.7 cm mark by a string attached to t

Question

A 0.102 kg meterstick is supported at its
37.7 cm mark by a string attached to the ceil-
ing. A 0.642 kg mass hangs vertically from the
4.88 cm mark. A mass is attached somewhere
on the meterstick to keep it horizontal and in
both rotational and translational equilibrium.
The force applied by the string attaching the
meter stick to the ceiling is 20 N.
Find the value of the unknown mass. The
acceleration of gravity is 9.81 m/s2 .
Answer in units of kg
Find the point where the mass attaches to the
stick.
Answer in units of cm

Explanation / Answer

First, we can work with the sum of the forces. The upward forces must equal the downward forces

The upward is only the 20 N force up

The downward will be the two attached masses and the mass of the meterstick

20 = (.102)(9.81) + (.642)(9.81) + (m)(9.81)

m = 1.295 kg

To find where to attached the mass, we can use the sum of the torques about a rotation point. If we use the rotation point at the zero location of the metersitck we get a torque equation of

(.642)(9.81)(4.88) + (.102)(9.81)(50) - (20)(37.7) + (1.295)(9.81)(x) = 0

x = 52.99 cm

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