A physical therapy patient supports his entire weight on crutches inclined outward at 25 degrees to the vertical. (a) Find the minimum coefficient of static friction between crutches and floor so that crutches won't slip.
Answer: (a) 0.466
How do I find this answer?
Answer: (a) 0.466
How do I find this answer?
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Halve the weight, as there are 2 crutches. You don't know weight, so assume 9.8N.
(9.8/2) = 4.9.
(4.9 x tan 25)/4.9 = 0.46631. Note multiplying tan 25 x 4.9, but then dividing it again by 4.9.
So really, the coefficient is just (tan 25), = 0.466.
(9.8/2) = 4.9.
(4.9 x tan 25)/4.9 = 0.46631. Note multiplying tan 25 x 4.9, but then dividing it again by 4.9.
So really, the coefficient is just (tan 25), = 0.466.
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at equilibrium must be
horizontal componenent of weight = friction force
mgsinθ=μmgcosθ
then
μ=tanθ
μ=tan25=0.466
horizontal componenent of weight = friction force
mgsinθ=μmgcosθ
then
μ=tanθ
μ=tan25=0.466