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larisa86 [58]
4 years ago
7

A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal.

Physics
1 answer:
Elanso [62]4 years ago
3 0

Answer:

we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.

Coefficient of static friction = 0.26

Explanation:

Given that:

length of the ladder = 16.0 m

weight of the ladder = 520 N

angle θ = 65.0°

(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :

Force = 850 N

distance of the climber from the base of the ladder = 4.20 m

The diagrammatic illustration representing what the given information entails can be seen from the attached file below.

Let consider the Ladder being at point A with the horizontal layer of the ground.

From the whole system; the condition for the equilibrium at the point A can be computed as :

N_2 (16  \  Sin\  65) = 850(4.2 \   \times Cos \  65 )+ 520 (\dfrac{16}{2}) Cos  \ 65

N_2 (14.50) = 850(1.7749 )+ 520 (8) \times 0.4226

N_2 (14.50) = 1508.665+1758.016

N_2 (14.50) = 3266.681

N_2  =\dfrac{ 3266.681}{14.50}

N_2  =225.28 \ N

N_1 = mg+F\\

where ;

w =mg

N_1 = 520+850

N_1 = 1370 \ N

Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.

(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:

N_2 (16  \  Sin\  65) = 850(9.4 \   \times Cos \  65 )+ 520 (\dfrac{16}{2}) Cos  \ 65

N_2 (14.50) = 850(3.9726 )+ 520 (8) \times 0.4226

N_2 (14.50) =3376.71+1758.016

N_2 (14.50) =5134.726

N_2  =\dfrac{5134.726}{14.50}

N_2  =354.12 \ N

Therefore; the coefficient of the static friction is;

\mu = \dfrac{f_s}{N_1}

\mu = \dfrac{354.12}{1370}

\mu  = 0.26

Coefficient of static friction = 0.26

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