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 58.0 kg. The rope makes an angle of θ = 25.0o with the vertical. Her legs are pushing out against the tree at an angle of ϕ = 19.5o above the horizontal. (a)What is tension in rope? (in N) (b)What must be the minimum coefficient of static friction μk to keep her shoes from slipping down the trunk of the tree?

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 58.0 kg. The rope makes an angle of 8 = 25.0° with the vertical. Her legs are pushing out against the tree at an angle of -19.5° above the horizontal (a) What is tension in rope? N (b) What must be the minimum coefficient of static friction to keep her shoes from slipping down the trunk of the tree?

Answer 1

T T cos T sine Biologist F sin F F Tree F coso f mg

Free body diagram for the forces acting on biologist is shown in figure given above.

Force mg is weight of biologist . Tension force T acting through the rope is resolved as ( T cos$\theta$ ) and         ( T sin$\theta$ ) as shown in figure. Reaction force F from tree that is acting on Biologist is also resolved as ( F cos$\phi$ ) and ( F sin$\phi$ ) . Friction force acting on biologist shoe is shown as f in the figure .

If we balance the forces in the vertical direction for biologist, we get

T cos$\theta$ = m g .................(1)

Hence tension force T = ( 58 $\times$ 9.8 ) / cos25 = 627 N

If we balance the forces in the horizontal direction for biologist, we get

T sin$\theta$ = F cos$\phi$ .................(2)

F = ( T sin$\theta$ ) / cos$\phi$ = ( 627 $\times$ sin25 ) / cos15 = 274 N .................... (3)

Friction force f should be equal to vertical component of Reaction force

Hence we get, f = $\mu$ F cos$\phi$ = F sin$\phi$ ......................(4)

where $\mu$ is friction coefficient

we get from eqn.(4) , $\mu$ = tan$\phi$ = tan(15) = 0.27