Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 12, Problem 13P

(a)

To determine

The horizontal and vertical forces the ground exerts on the base of the ladder.

(a)

Expert Solution
Check Mark

Answer to Problem 13P

The forces in the horizontal direction are, normal force exerts by the wall is 268N_, the static friction force is 268N_ and force in the vertical direction is, normal force exerted by the ground on the ladder is 1300N_.

Explanation of Solution

Write the first condition for equilibrium, the magnitude of the net external force on the object must be equal zero:

  Fext=0                                                                                                       (I)

Here, Fext is algebraic sum of all forces acting on the object in the x and y -directions.

Write the second condition for equilibrium, the magnitude of net external torque on the object about a fixpoint must be zero:

  τext=0                                                                                                      (II)

Here, τext is algebraic sum of all torques acting on the object about any point.

The forces acting on the ladder are shown as.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 12, Problem 13P

Write the first equilibrium condition for the ladder in x -direction as

  fsnw=0                                                                                                   (III)

Here, fs is the force of static friction, nw normal force exerts by the wall.

Write the first equilibrium condition for the ladder in y -direction as

  ngm1gm2g=0                                                                                     (IV)

Here, ng is the normal force exerted by the ground on the ladder, m1 is the mass of the ladder, m2 is mass of the firefighter and g is gravitational acceleration.

Write the expression for the weight of the body.

  F=mg                                                                                                        (V)

Here, m is the corresponding mass of the body.

If the mass is m1 then the force becomes F1 or if the mass is m2 then the force become F2.

Write the expression for second equilibrium for the ladder in as

  nw(CF)m1g(DO)m2g(EO)=0                                                         (VI)

Here, nw normal force exerts by the wall, CF distance between the line of action of force nw to the ground, DO is distance between the line of action of force F1 or m1g to the point and O, EO distance between line of action of force F2 of m2g to the point O.

Substitute Lsinθ for CF, L2cosθ for DO and xcosθ for EO in equation (IV).

  nw(Lsinθ)m1g(L2cosθ)m2g(xcosθ)=0

Here, L is the length of the ladder and θ is the angle ladder makes with the horizontal.

Simplify the above expression for nw as.

  nw(Lsinθ)=m1g(L2cosθ)+m2g(xcosθ)

Re-arrange the terms.

  nw=gcosθ(m1L2+xm2)Lsinθ

Substitute gcosθ(m1L2+xm2)Lsinθ for nw in equation (I).

  fsgcosθ(m1L2+xm2)Lsinθ=0

Re-arrange the terms.

  fs=cosθ(m1gL2+xm2g)Lsinθ                                                                           (V)

Conclusion:

Substitute 9.81m/s2 for g, 60° for θ, 500N for m1g, 800N for m2g, 15m for L and 4m for x in equation (V).

  fs=cos60°((500N)(15m)2+(4m)(800N))(15m)sin60°=267.5056N268N

Substitute 268N for fs in equation (III).

  (268N)nw=0

Simplify the above expression for nw as.

  nw=268N

Substitute 500N for m1g, 800N for m2g, in equation (II)

  ng(500N)(800N)=0

Simplify the above expression for ng as.

  ng=1300N

Thus, the forces in horizontal direction are, normal force exerts by the wall is 268N_,static friction force is 268N_ and force in vertical direction is, normal force exerted by the ground on the ladder is 1300N_.

(b)

To determine

The coefficient of static friction at the verge of sliding between ladder and ground.

(b)

Expert Solution
Check Mark

Answer to Problem 13P

The coefficient of static friction at the verge of sliding between ladder and ground is 0.324_.

Explanation of Solution

At the verge of slipping the force of friction (fs) must have its maximum value as.

  (fs)max=μsng                                                                                             (VI)

Here, μs is maximum static friction coefficient and (fs)max is maximum friction force.

At the verge of slipping, substitute x for x in equation (V).

  (fs)max=gcosθ(m1L2+xm2)Lsinθ                                                                (VII)

Here, x is the new position of the fire fighter.

Conclusion:

Substitute gcosθ(m1L2+xm2)Lsinθ for (fs)max and g(m1+m2) for ng in equation (VII).

  μs(g(m1+m2))=gcosθ(m1L2+xm2)Lsinθ

Re-arrange the terms,

  μs=cosθ(m1gL2+xm2g)L(m1g+m2g)sinθ                                                                     (VIII)

Substitute 9.81m/s2 for g, 60° for θ, 500N for m1g, 800N for m2g, 15.0m for L and 9.00m for x in equation (VIII).

  μs=cos60°((500N)(15.0m)2+(9.00m)(800N))((500N)+(800N))(15.0m)sin60°=0.324

Thus, the coefficient of static friction at the verge of sliding between ladder and ground is 0.324_.

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Chapter 12 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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