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question 19 of chapter 10 rotation and rolling...///////////////////////////////

question 19 of chapter 10 rotation and rolling...///////////////////////////////

Grade:11

1 Answers

we will so
101 Points
6 years ago
A force of 6 N acting at an angle of 30° is just able to loosen the wrench at a distance 8 cm from it.
Therefore total torque acting at A about the point 0
= 6 sin 30° × (8/100)
Therefore total torque required at B about the point 0
= F × 16/100  F × 16/100 = 6 sin 30° × 8/100
 F = (8 × 3) / 16 = 1.5 N.

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