sec_phys_forces_problems

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University of British Columbia *

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PHY 159

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Physics

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Jan 9, 2024

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Physics Forces Problems Science and Mathematics Education Research Group Supported by UBC Teaching and Learning Enhancement Fund 2012-2015 FACULTY OF EDUCATION FACULTY OF EDUCATION Department of Curriculum and Pedagogy F A C U L T Y O F E D U C A T I O N

Question Title Question Title Forces Problems Retrieved from: http://www.technologytom.com/html/forces_in_bridges.html

Question Title Question Title Forces Problems The following questions have been compiled from a collection of questions submitted on PeerWise ( https://peerwise.cs.auckland.ac.nz/ ) by teacher candidates as part of the EDCP 357 physics methods courses at UBC.

Question Title Question Title Forces Problems I Bob and Yoshiko are having a tug-of-war on the wooden floor in the gymnasium. Bob has a mass of 78.5 kg and Yoshiko weighs 980 N. If Bob pulls with a force of 75N and is winning the tug-of-war (pulling Yoshiko towards him), what force F Y is Yoshiko pulling with on the rope? Assume the acceleration due to gravity is 9.81 m/s 2 . A. F Y = 55 N B. F Y = 70 N C. F Y = 75 N D. F Y = 80 N

Question Title Question Title Solution Answer: C Justification: No matter who is winning the tug-of-war, the forces applied on either end of the rope must be equal in magnitude, but opposite in direction or else the rope will break or slip out of one of their hands. Therefore the force applied by Yoshiko on the rope is F Y = 75 N (Answer C ). Bob Yoshiko F B F Y

Question Title Question Title Solution continued Further Explanation: The person who is "winning the tug-of- war“ here is the person who has the greatest force of friction working against them. As Bob pulls on the rope he pushes the ground towards Yoshiko and as a result, the force of friction on Bob acts in the direction opposite of Yoshiko. A similar argument can be made for the force of friction on Yoshiko that pulls Bob. However, for Bob to be "winning" the force of friction being applied to him by the ground must be greater than the force of friction being applied to Yoshiko by the ground. Furthermore, this is only possible if the coefficient of static friction for Bob (μ B ) is larger than the coefficient of static friction for Yoshiko (μ Y ) because Yoshiko has a larger mass and therefore if both Bob and Yoshiko had the same coefficient of static friction between them and the floor Yoshiko will always win because the force of friction F f = μm g .

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