Question

A biologist inspects an owl's nest that is built in a hollow cavity of a tree. Refer to the image. The mass of the biologist is 47.0 kg. The rope makes an angle of θ = 26.5^o with the vertical. Her legs are pushing out against the tree at an angle of ϕ = 19.5^o above the horizontal.

0 T E.

(a)

What is tension in rope? N

(b)

What must be the minimum coefficient of static friction μ_k to keep her shoes from slipping down the trunk of the tree?

Answer 1

Resolving T in to two components,

T cos26.5^o vertically upward and

T sin26.5^o towards the tree and perpendicular to it

Similarly, resolving F in to two components,

F cos19.5^o, away from tree, perpendicular to it, which is the normal force

F sin19.5^o, vertically upward, along the tree, which provides the friction

Weight mg of the biologist act vertically downward

Now, for translational equilibrium in horizontal direction,

F cos19.5^o = T sin26.5^o

Or, F = T(sin26.5^o/cos19.5^o)

= 0.47334T

In vertical direction, equilibrium needs,

F sin19.5^o + T cos26.5^o = mg

Or, 0.47334T*0.3338 + T*0.89493 = 47*9.8

Or, 1.0529*T = 460.6

Thus, tension, T = (460.6/1.0529) = 437.46 N

Coefficient of friction is given by,

$\mu$ _k = Frictional force/Normal force

= (F sin19.5^o/F cos19.5^o)

= tan 19.5^o

= 0.35