A diver of mass 60kg stands on the end of a diving board of length 2m from the pivot point. Calculate the upward force exerted on the retaining spring 30cm from the pivot point.

This question is asking about forces acting around a pivot point, so we can use the formula for the moment M = F*D about the pivot point. The clockwise and anticlockwise moments around the pivot must be equal, otherwise the diving board would start rotating. The clockwise moment would be the weight of the man, his mass * g , roughly equal to 600 N (if g is about 10), multiplied by the distance from the pivot point, 2m. This equals 1200 Nm. This balances the anticlockwise moment, which is the distance from the pivot point, 0.3 m, multiplied by the upwards force (tension). We find the force by dividing both sides by 0.3m, to determine an answer of 4000 N.

LG
Answered by Louis G. Physics tutor

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