Problem 11.35 An 71kg person stands on a uniform 7.4kg ladder that is 40 m long,

Question

Problem 11.35 An 71kg person stands on a uniform 7.4kg ladder that is 40 m long, as shown in the figure(Figure 1) The floor is rough: hence, it exerts both a normal force, f1, and a frictional force, f2, on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f3. Part A Using the dimensions given in the figure, find the magnitudes of f1, f2, and f3. Express your answer using two significant figures. f1= kN Submit MY Answers Give Up [ Incorrect; Try Again; 3 attempts remaining ] Part B Express your answer using two significant figures. f2= kN

Explanation / Answer

Length of the ladder L = 4 m

Angle = sin ^ -1 (3.8 / 4 ) = 71.8 degrees

position of the ladder d = b / cos 71.8

= (0.70 m) / cos 71.8

=2.24 m

Balancing vertical forces,
f? = m? + m?

When the ladder is on the verge of slipping, frictional force from right to left at the bottom of the ladder with the floor,
F = ?_min * f? = ?_min * (m? + m?)

Taking moments about the point of contact of the ladder with the wall,
f? * Lcos? - F * Lsin? - m? * (L - d) cos? - m? * (L/2) cos? = 0
=> (m?+m?) * Lcos? - ?_min * (m?+m?) * Lsin? - m? * (L - d) cos? - m? * (L/2) cos? = 0
=> ?_min * (m?+m?) * Lsin? = (m?+m?) * Lcos? - m? * (L - d) cos? - m? * (L/2) cos?
=> ?_min * (m?+m?) * Lsin? = m? * dcos? + (m?/2) Lcos?

=> ?_min
= (dm? + Lm?/2)cot? / [L(m?+m?)]
= [(d/L)m? + (1/2)m?] cot? / (m?+m?).

2)
Balancing vertical forces,
f? = (m? + m? ) g

= ( 71 kg ) + ( 7.4 kg ) (9.8 m/s^2) = 768.32 N
Balancing horizontal forces,
f1 = f3
Taking moments of all the forces about the point of contact of the ladder with the floor,
- N? Lsin? + m?g (L/2)cos? + m?g dcos? = 0
f3 = f? = (m?/2 + m?d/L) gcot?

= (7.4 kg /2 + ( 71 kg ) (2.24 m) / 4 ) ( 9.8 m/s^2) cot 71.8

=151.9 N

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