A0.117 kg meterstick is supported at its 37.2 cm mark by a string attached to th

Question

A0.117 kg meterstick is supported at its

37.2 cm mark by a string attached to theceil-

ing. A0.694 kg mass hangs vertically from the

4.49 cm mark. A mass is attached somewhere

on themeterstick to keep it horizontal and in

both rotational andtranslational equilibrium.

The force applied by the string attachingthe

meterstick to the ceiling is 22.2 N.

The acceleration of gravity is 9.81 m/s2 .

a) Find the value of the unknownmass.

Answer in units of kg.

006 (part 2 of 2) 10.0points

b) Find the point where the mass attachesto

thestick. Answer in units of m.

Explanation / Answer

Mass of meterstick plus weights = 22.2 N/9.81 m/s^2 = 2.26kg Let the meterstick lie such that the smaller numbers are tothe left Part a. Unknown mass = 2.26 - 0.117 -0.694 = 1.45kg Part b. Let x be the distance in cm to the right of37.2 which is the pivot point Then x * 1.45 = (37.2-4.49)*0.694 or x = 15.66 or the hanging point is 37.2 + 15.66 = ...cm or x = 15.66 or the hanging point is 37.2 + 15.66 = ...cm

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