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What is the weight of the measuring stick if it is balanced by a support force at the 1 m mark? The acceleration of gravity is 9.81 m/s 2 . Answer in units of n.

User Alcaprar
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1 Answer

2 votes

Answer:

7.85 N

Step-by-step explanation:

The measuring stick is uniform so we can safely assume that all of its mass is concentrated at the center.

The center is at 7/2 i.e 3.5 meter mark.

Given that it is balanced, the sum of clockwise force is equal to the sum of clockwise force according to the principle of moments. The 1 meter mark is the pivotal point.

Torque = Force * distance(perpendicular to the force);

It is balanced so:

Torque due to 2 kg rock = torque due to weight of the measuring stick;

2 * 9.81 * 1 = x *9.81 * (3.5-1); x is the mass of the measuring stick;

x= 0.8 kilograms

Weight = 0.8 *9.81 =7.85 N

User Antonio Val
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