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Problem 13

A worker pushes horizontally on a 35 $\mathrm{kg}…

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Problem 12

In about $1915,$ Henry sincosky of Philadelphia suspended himself from a rafter by gripping the rafter with the thumb of each hand on one side and the fingers on the opposite side (Fig. $6-21 )$ . Sincosky's mass was 79 kg. If the coefficient of static friction between hand and rafter was $0.70,$ what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers? (After suspending himself, Sincosky chinned himself on the rafter and then moved hand-over-hand along the rafter. If you do not think Sincosky's grip was remarkable, try to repeat his stunt.)

Answer

$F_{N}=280 N$


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Video Transcript

so here there is no acceleration. So the upward static friction forces there being four of them, one for each thumb and one for each set of opposing fingers essentially equals the magnitude of the downward pull of gravity. So we can say that the force of friction static times four would be equal to M G. Or we can say four times the coefficient of static friction. Times forced normal would be equal to M G. Uh, this would be equal to 79 kilograms. Most played by 9.80 meters per second squared. And so we can say that if the coefficient of static friction equals 0.70 on the force, normal would then be equal to 79 kilograms multiplied by 9.80 meters per second squared divided by four times 40.70 And we find that the force normal here would be approximately 280 Nunes to react around to two significant figures. 218 unions would be the approximate value of the forces of the force normal. That is the end of the solution. Thank you for watching

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