A sphere with a mass of 9 kg and radius of 1.7 meters is being pressed intoa ste

Question

A sphere with a mass of 9 kg and radius of 1.7 meters is being pressed intoa step with a height of 0.3 meters (see sketch). Because the corner of the step contacts the sphere, it has an x and y component to its normal force on the sphere as shown. A force is applied to the sphere in the +x direction as indicated. What does the minimum value this force needs to be such that normal force between the sphere and the horizontal surface is zero and the sphere starts to lift over the step? Answer in Newtons.

Explanation / Answer

Writing torque equation about the corner of the step,.

F(r-h) = mg*d

Now r-h = 1.7-0.3 = 1.4m

d =sqrt(1.7^2-1.4^2) = 0.9644m

So, 1.4F = 9*9.8*0.9644

F = 9*9.8*0.9644/1.4

= 60.76 N answer

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