Frictional forces on mobility walkers
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Frictional forces on mobility walkers
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[post_date] => 2025-01-09 07:34:10
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[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
A mobility walker or “walker” is useful for people who are at risk of falling. This includes the elderly, individuals with severe knee arthritis, and patients recovering from accidents or invasive leg surgery. Instead of maintaining the typical upright position, a walker allows the person to transmit some of their body weight from their legs, through their arms, and onto the walker by leaning on it.
To reduce friction, people often place tennis balls on the bottom feet of the walker. Compared to the rubber grip originally attached to the walker, the tennis balls have a significantly lower coefficient of friction. This allows the walker to slide across the ground with much less applied force by the patient, making it further useful for those with additional weakness in the arms or upper torso.
The diagram depicts an example of a 2.0 kg mobility walker being slid across the floor due to an applied force of 20N directed at an angle of 60°. Assume the walker is sliding across the floor with constant velocity.
Figure 1. A 20N force applied to a walker with tennis balls attached to the bottom.
[post_title] => Frictional forces on mobility walkers
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[question] => While the 20N force is applied, how will the normal force exerted on the walker upwards from the ground compare to the weight of the walker?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
Weight is defined to be kg.
The force is pushing the walker down into the ground.
Since the walker is being pushed into the ground, the normal force from the floor will have to be larger than the weight of the walker to maintain ΣF = 0 in the vertical direction.
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[answers] => Array
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[each_answer] => A. The magnitude of the normal force will be larger than the weight of the walker.
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[1] => Array
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[each_answer] => B. There is no normal force since the walker is supported by the floor.
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[2] => Array
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[each_answer] => C. The magnitude of the normal force will be smaller than the weight of the walker.
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[3] => Array
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[each_answer] => D. The magnitude of the normal force will be the same as the weight of the walker.
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => What is the magnitude of the normal force exerted on the walker from the ground while the 20N force is applied?
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[answer] => 2
[description] => Reason for the Correct Answer:
The walker has no vertical acceleration.
Use ΣF = 0 for the vertical direction.
F(normal) – F(sinθ) – mg = 0.
F(normal) = 20N(sin60°) + 2.0kg(9.8m/s²)
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[answers] => Array
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[0] => Array
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[each_answer] => A. 20Ncos(60) * 2.0(9.8m/s²)
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[1] => Array
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[each_answer] => B. 20Nsin(60) + 2.0(9.8m/s²)
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[2] => Array
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[each_answer] => C. 20Nsin(60) – 2.0(9.8m/s²)
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[each_answer] => D. 20Ncos(60) / 2.0(9.8m/s²)
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => What is the size of the frictional force on the mobility walker while the 20N force is applied?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
The walker is sliding across the floor with constant velocity.
Use ΣF = 0 for the horizontal direction. The only forces in the horizontal direction are friction backwards and the horizontal component of the 20N force.
F(friction) – F(cosθ) = 0
F(friction) – 20N(cos60°) = 0, so F(friction) = 20Ncos60
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[answers] => Array
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[0] => Array
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[each_answer] => A. 20Ncos60 – 2.0(9.8m/s²)
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[1] => Array
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[each_answer] => B. 20Nsin60
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[2] => Array
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[each_answer] => C. 20Ncos60
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[3] => Array
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[each_answer] => D. 20Nsin60 + 2.0(9.8m/s²)
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[3] => Array
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[quiz_unique_key] => 2261298308
[question] => If the person changes the angle of the 20N force from 60 degrees to 30 degrees, how will the normal force change?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
Think about ΣF = 0 for the vertical direction.
As θ decreases, sin(θ) decreases.
Since the vertical component of the 20N force is decreasing, the normal force decreases.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The normal force will increase.
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[1] => Array
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[each_answer] => B. The normal force will increase first, then decrease.
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[each_answer] => C. The normal force will decrease.
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[each_answer] => D. The normal force will stay the same.
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[question] => The person finishes using the walker and lets it sit at rest on the ground. While the walker is sitting at rest, what is the normal force exerted on the walker by the ground?
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[answer] => 3
[description] => Reason for the Correct Answer:
Think about ΣF = 0 for the vertical direction.
While the walker is sitting at rest, the only forces are the normal force and the force of gravity.
ΣF = 0, F(normal) – mg =0, F(normal) = mg = (2.0kg)(9.8m/s²) = 19.6N
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[answers] => Array
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[each_answer] => A. 4.9 N
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[each_answer] => B. 0 N since no one is pushing down on it
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[2] => Array
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[each_answer] => C. 19.6 N
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[each_answer] => D. 9.8 N
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