College Physics
College Physics
5th Edition
ISBN: 9781260486841
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 4, Problem 72P

(a)

To determine

Maximum and minimum upward force that the supporting cables exert on the elevator car.

(a)

Expert Solution
Check Mark

Answer to Problem 72P

Maximum force is 26400N_

Minimum force is 20640N_

Explanation of Solution

Write the equation of force exerted by the supporting cables on the elevator car.

    F=ma        (I)

Here m is the mass of the elevator and a is the acceleration.

Write the equation of force due to gravity

    F=mg        (II)

Here g is the acceleration due to gravity.

Then the maximum force would be

    Fmax=mg+ma=m(g+a)        (III)

The minimum force would be

    Fmin=m(ga)        (IV)

Conclusion

Substitute 2400kg for m, 1.2m/s2 for a and 9.8m/s2 for g in (III)

    Fmax=2400kg(9.8m/s2+1.2m/s2)=26400N

Substitute 2400kg for m, 1.2m/s2 for a and 9.8m/s2 for g in (IV)

  Fmin=2400kg(9.8m/s21.2m/s2)=20640N

Maximum force is 26400N_.

Minimum force is 20640N_.

(b)

To determine

Minimum time to ascent from the lobby to the observation deck.

(b)

Expert Solution
Check Mark

Answer to Problem 72P

The time is 102s_.

Explanation of Solution

Write the expression to find the time taken to reach the maximum ascent speed.

    v=u+at1        (I)

Here, v is the maximum ascent speed, u is the initial speed, a is the acceleration, and t1 is the time taken to reach the maximum ascent speed.

Write the expression to find the distance covered during acceleration.

    s1=ut+12at12

Here, s1 is the vertical displacement

Write the expression to find the distance covered.

    s2=vt2

Here, s2 is the vertical displacement and t2 is the time of displacement with uniform speed.

Conclusion:

Substitute 1.2m/s2 for a, 18m/s for v and 0m/s for u in (I)

    18m/s=0+(1.2m/s2)t1t1=18m/s1.2m/s2=15s

Vertical distance covered during this time is,

Substitute 1.2m/s2 for a, 15s for t1 and 0m/s for u in (I)

    s1=0(15s)+12(1.2m/s2)(15s)2=135m

Remaining vertical distance is,

    s2=640m135m=505m

Elevator moves with uniform speed to cover this distance.

Expression to calculate the time taken to cover this distance,

    t2=s2v

Substitute 505m and 18m/s for v.

    t2=505m18m/s=28s

The minimum time taken by the elevator to ascend from the lobby to the observation desk is,

    t=t1+t2

Substitute 15s for t1 and 28s for t2 to find t.

    t=15s+28s=43s

Therefore, the minimum time taken by the elevator to ascend from the lobby to the observation desk is 43s_.

(c)

To determine

Maximum and minimum values of a passenger’s apparent weight during the ascend.

(c)

Expert Solution
Check Mark

Answer to Problem 72P

The maximum weight during the ascent is 660N_.

The minimum weight during the ascend is 516N_

Explanation of Solution

The maximum weight would be

    W=M(g+a)        (IX)

Here W is the weight and M is the mass of the passenger

The minimum weight would be

    W=M(ga)        (X)

Conclusion

Substitute 60kg for M, 1.2m/s2 for a and 9.8m/s2 for g in (IX)

    Fmax=60kg(9.8m/s2+1.2m/s2)=660N

Substitute 60kg for M, 1.2m/s2 for a and 9.8m/s2 for g in (X)

    W=60kg(9.8m/s21.2m/s2)=516N

The maximum weight during the ascent is 660N_.

The minimum weight during the ascend is 516N_

(d)

To determine

Minimum time for the elevator to descend to the lobby.

(d)

Expert Solution
Check Mark

Answer to Problem 72P

The time is 12.49s_

Explanation of Solution

Write the expression to find the time taken to reach the maximum decent speed.

    v=u+at1        (II)

Here, v is the maximum ascent speed, u is the initial speed, a is the acceleration, and t1 is the time taken to reach the maximum descent speed.

Write the expression to find the distance covered during acceleration.

    s1=ut+12at12

Here, s1 is the vertical downward displacement

Write the expression to find the distance covered.

    s2=vt2

Here, s2 is the vertical downward displacement and t2 is the time of displacement with uniform speed.

Conclusion

Substitute 1.2m/s2 for a, 10m/s for v and 0m/s for u in (I)

    10m/s=0+(1.2m/s2)t1t1=10m/s1.2m/s2=8.33s

Vertical distance covered during this time is,

Substitute 1.2m/s2 for a, 8.33s for t1 and 0m/s for u in (I)

    s1=0(8.33s)+12(1.2m/s2)(8.33s)2=41.6m

Remaining vertical distance is,

    s2=640m41.6m=598.4m

Elevator moves with uniform speed to cover this distance.

Expression to calculate the time taken to cover this distance,

    t2=s2v

Substitute 598.4m and 10m/s for v.

    t2=598.4m10m/s=59.84s

The minimum time taken by the elevator to ascend from the lobby to the observation desk is,

    t=t1+t2

Substitute 8.33s for t1 and 59.84s for t2 to find t.

    t=8.33s+59.84s=68.2s68s

Therefore, the minimum time taken by the elevator to ascend from the lobby to the observation desk is 68s.

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Chapter 4 Solutions

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