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 other side (see the figure). Sincosky's mass was 87.6 kg. If the coefficient of static friction between hand and rafter was 0.743, what was the least magnitude of the normal force on the rafter from each thumb or opposite fingers?
I thought that k*normal >= weight, where k is the coefficient of friction so the normal would equal (87.6*9.8)/0.743
Nevermind, the problem was that I had to divide the resultant by 4
To find the least magnitude of the normal force on the rafter from each thumb or opposite fingers, you can indeed use the inequality you mentioned:
k * normal ≥ weight
In this case, the weight refers to the gravitational force acting on Henry Sincosky, which can be calculated as:
weight = mass * acceleration due to gravity
So, weight = 87.6 kg * 9.8 m/s^2 = 857.28 N
Next, you can rearrange the inequality to solve for the normal force:
normal ≥ weight / k
Plugging in the values, you would get:
normal ≥ 857.28 N / 0.743
Now you can solve for the least magnitude of the normal force.