Question

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 of the other side. Sincosky's mass was 79 kg. If the coefficient of static friction between hand and rafter was 0.52, 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.)

Fnormal, min = N

Answer 1

• 372.2 N

Step-by-step explanation:

mass (m) = 79 kg

coefficient of static frictin (μ) = 0.52

acceleration due to gravity (g) = 9.8 m/s^{2}

• for Henry to suspend himself, it means the frictional force holding him up must be equal to his weight, Fs = W
• For each hand he grips a side of the rafter with his thumb and the other side with the remaining fingers, which is 2 contact points per hand and 4 contact surfaces for both hands. Hence 4Fs = W
• 4Fs = 4μFn and W = mg hence 4μFn = mg and Fn = mg/4μ

• Fn = mg/4μ =
  `(79 x 9.8)/(4 x 0.52)` = 372.2 N