# Someone of mass 100 kg is standing on top of a vertical cliff with only an old rope that he knows will support no more than 500 N. His plan is to slide.

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Someone of mass 100 kg is standing on top of a vertical cliff with only an old rope that he knows will support no more than 500 N. His plan is to slide down the rope using friction to keep from falling freely. At what minimum rate can he accelerate down the rope in order not to break it?

Newton’s first law can not be applied to problems in which
Reading Assignment 5.1: Newton’s first law can not be applied to problems in which Objects are at rest. Only one force acts on an object. Many forces combine to keep an object in position. An object is sliding along at constant speed. L13sp09 2

The normal force is present in every situation.
Reading Assignment 5.1/2: Which of the following statements is true? The normal force is present in every situation. The normal force is always upward and equal to mg. The normal force occurs only if the object is in contact with a surface. An object can have only one normal force.

Practicing free-body diagrams, components, analysis
A person pushes a 40-kg shopping cart with a force of 100 N acting down at 45. The cart travels at a constant speed of 1.0 m/s. What is the value of the combined retarding force acting horizontally on the cart due to friction and air drag? How much force does the cart exert on the ground?

Practicing free-body diagrams, components, analysis
The same person with shopping cart is going up a 15o-parking ramp. The cart travels at a constant speed of 1.0 m/s. How much force does the person need to exert in order to maintain the speed? How much force does the cart exert on the ground? 15o

Practicing free-body diagrams, components, analysis
A lifting beam at an upper barn window is used to lift 60-kg packs of hey into the loft of the barn. How much force does the person exert to lift the pack at constant speed?

Practicing free-body diagrams, components, analysis
How much force does the person exert to lift the pack at constant speed with this modified pulley scheme?

Practicing free-body diagrams, components, analysis
At the bottom, the person must bring the speed of the pack from 0 to 1.0 m/s. How much tension will the rope incur, if this speed change happens within 0.5 s?

Practicing free-body diagrams, components, analysis
A 40-kg child in a 20-kg dumb-waiter is pulling himself upward. How much force does the child need to exert in order to achieve a minimum upward acceleration of 0.1 m/s2? How much is the contact force between box and child?

Apparent weight = normal force with level ground
A 60-kg person is using an elevator. What would a scale read for the following parts of the trip? N N – mg = 0 mg N = mg = 590 N The elevator is resting

Apparent weight = normal force with level ground
A 60-kg person is using an elevator. What would a scale read for the following parts of the trip? N – mg = ma N N = m (a+g) mg N = 60 kg ( )m/s2 N = 710 N Acc. To 2 m/s within 1 sec The elevator is resting N – mg = 0 N = mg = 590 N

Apparent weight = normal force with level ground
A 60-kg person is using an elevator. What would a scale read for the following parts of the trip? N N – mg = 0 N = mg = 590 N mg Constant speed 2 m/s Acc. To 2 m/s within 1 sec N – mg = ma N = m (a+g) = 710 N The elevator is resting N – mg = 0 N = mg = 590 N

Apparent weight = normal force with level ground
A 60-kg person is using an elevator. What would a scale read for the following parts of the trip? N Slowing to zero in 1 sec N – mg = - ma mg N = m (g-a) = 60kg (9.8 – 2)m/s2 N = 470 N Constant speed 2 m/s N – mg = 0 N = mg = 590 N Acc. To 2 m/s within 1 sec N – mg = ma N = m (a+g) = 710 N The elevator is resting N – mg = 0 N = mg = 590 N

Apparent weight = normal force with level ground
A 60-kg person is using an elevator. What would a scale read for the following parts of the trip? N = 470 N Slowing to zero in 1 sec N – mg = - ma Going down at 2 m/s2 N – mg = - ma N = 470 N Constant speed 2 m/s N – mg = 0 N = mg = 590 N Acc. To 2 m/s within 1 sec N – mg = ma N = m (a+g) = 710 N The elevator is resting N – mg = 0 N = mg = 590 N

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