Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 12, Problem 67P

(a)

To determine

To Calculate: The force that exerted by the ladder against the wall.

(a)

Expert Solution
Check Mark

Answer to Problem 67P

  FBythewall=0.15kN

Explanation of Solution

Given information :

Mass of the ladder =20.0kg

Tension of the wire =29.4N

Breaking point tension of the wire =200N

Mass of the person 80.0kg

Formula used:

Torque can be obtained:

  τ=rF

Where, τ is the torque, F is the applied force and r is the distance between the point where force is applied to the rotational axis.

Calculation:

  Physics for Scientists and Engineers, Chapter 12, Problem 67P

The free-body diagram of the forces acting on the ladder.

Here,

  l= Length of the ladder

  m is the mass of the ladder and M is the mass of the person.

Apply τ=0 about an axis through the bottomof the ladder.

  Fbywall(lsinθ)mg(12lcosθ)Mg(12lcosθ)=0

Solve for Fbywall wall and simplify,

  FBythewall=(m+M)gcosθ2sinθFBythewall=(m+M)g2tanθ

From the figure given above determine θ .

  θ=tan1(5.0m1.5m)θ=73.3

  FBythewall=(20kg+80kg)(9.81m/s2)2tan73.3FBythewall=0.15kN

Conclusion:

The force that exerted by the ladder against the wall.

  FBythewall=0.15kN

(b)

To determine

To Calculate: The distance thatfrom the bottom end of the ladder that can an 80.0kg person climb.

(b)

Expert Solution
Check Mark

Answer to Problem 67P

  L=3.8m

Explanation of Solution

Given information :

Mass of the ladder =20.0kg

Tension of the wire =29.4N

Breaking point tension of the wire =200N

Mass of the person 80.0kg

Formula used :

From Newton’s second law of motion:

  F=ma

Where, m is the mass and a is the acceleration.

Calculation:

Apply Fx=0 to the ladder,

  TFBywall=0T=FBywall

Apply τ=0 about an axis through the bottom of the ladder subject to the condition that FBywall=TMax :

  TMax(lsinθ)mg(12lcosθ)Mg(Lcosθ)=0

Here, L is the maximum distance along the ladder that the person can climb without exceeding the maximum tension in the wire.

  L=TMaxlsinθ12mglcosθMgcosθ

  L=(200N)(5.0m)12(20kg)(9.81m/s2)(1.5m)(80kg)(9.81m/s2)cos73.3L=3.8m

Conclusion:

The distance thatfrom the bottom end of the ladder that can an 80.0kg person climb is L=3.8m.

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