Question

A Motor cyclist at a speed of 5m/s is describing the circle with radius 25cm. Find his inclination with vertical. what is the value of coefficient of friction between tyre of ground ​

Answer 1

Explanation:

A uniform metre scale has two weights of 10 gf and 8 gf suspended at the 10 cm and 80 cm marks respectively. If the metre scale itself weights 50 gf, find where must the weight be, so that the metre scale stays balanced?

December 30, 2019Archa Tawari

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ANSWER

Let the meter scales is balanced at X cm mark on the scale.

At the balancing condition, Anticlock wise moment must be equal to clock wise moment.

Clockwise moment, about the balancing point is

= moment by 80 gf

=80gf×(80−X)cm

=6400gfcm−80Xgfcm.....(1)

Anticlock wise moment about the balancing point is

= moment by 10 gf + moment by 50 gf

=10gf×(X−10)+50gf×(X−50)

=10Xgfc,−100gfcm+50Xgfcm−2500gfcm.....(2)

At the balancing condition, Clock wise at equilibrium = Anti clock wise moments.

6400−80X=100X−100+50X−2500

6400+2600=140X⇒140X=8000

⇒X=1408000=64.28

Scale is balanced at 64.28 cm from the beginning.