A uniform 240-g meter stick can be balanced by a 240-g weight placed at the 100-

Question

A uniform 240-g meter stick can be balanced by a 240-g weight placed at the 100-cm mark if the fulcrum is placed at the point marked: 75 cm 70 cm 65 cm 60 cm 55 cm Calculate the torque (magnitude and direction) about the post A of a diving board, exerted by a 54-kg person who is 3 meters from the supporting post B: 530 N m, clockwise. 1590 N m, clockwise. 1590 N m, counterclockwise. 2120 N m, counterclockwise. 2120 N m, clockwise. An object at the surface of Earth (at a distance R from the center of Earth) weighs 90 N. Its weight at a distance 3R from the center of Earth is 10N 30 N 90 N 270 N 810 N

Explanation / Answer

9)

Answer :A (75cm)

Given that

The mass of the meter stick is (M) =240g =240*10-3kg

The mass of weight balanced is m =240g=240*10-3kg

Now the mass of the stics a the center of the stick that at 50cm mark

0cm 50cm 10cm

Mg <x1...... ..A. .........x2 mg

Now applying the torque about the point A is given by

Mgx1-mgx2 =0

Mgx1 =mgx2

x1 =x2

Now the fulcrum must be placed at the center of 100cm mark and 50 cm mark that is at 75cm mark

11)

Answer :A (10N)

We know that

Acceleration due to gravity g =GMe/r2

When it the case of weight i.e mg =GMem/r2

At a distance R then 90N =GMem/R2

At a distance 3R then mg =GMem/(3R)2

=GMem/9R2

Then from the above we can say that at distance of 3R the weight becomes 90N/9 =10N

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