What is the weight of the motor?Two people carry a heavy electric motor by placing it on a light board 2.70 m long. One person lifts at one endwith a force of 400N. and the other lifts the opposite end with a force of 850N.w = 1050 NYou may want to review (Page)For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Locating yourcenter of gravity while you work outSubmitPrevloun Anawerev CorrectIDENTIFY: Apply EF, = 0 and Er,= 0 to the board.SET UP: Let +y be upward.EXECUTE: If the board is taken to be massless, the weight of the motor is the sum of the applied forces, 1050 N.Part BWhere along the board is its center of gravity located?z= 1.67 m from the end where the 400-N force is appliedSubmitPrevlous Anawerev CorrectSET UP: Let z be the distance of the center of gravity of the motor from the end of the board where the 400-N force is applied.(2.70 m)(650 N)(1060 N)EXECUTE: The motor is a distance1.67 m from the end where the 400-N force is applied, and so is 1.03 m from the end where the 650-N force is applied.Part CSuppose the board is not light but weighs 220 N. with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case?Templates Symbols undo redo feset keyboard shortcuts HelpSubmitPrevlous Answere Requeet AnawerX Incorrect; Try Again; 3 attempts remainingPart DWhere is its center of gravity located?Templates Symbols undo redo Teset keyboard shortcuts helpm from the end where the 400-N force is applied

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Part C Suppose the board is not light but weighs 220 N. with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case? Temglates Symbols undo rado Tese keyboard shortcuts Help Submit Previous Answere Raqueet Answer x Incorrect; Try Again; 3 attempts remaining Part D Where is its center of gravity located? Templates Symbols undo rego Tese keyboard shortouts help m from the end where the 400-N force is applied

Part C Suppose the board is not light but weighs 220 N. with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case? Temglates Symbols undo rado Tese keyboard shortcuts Help Submit Previous Answere Raqueet Answer x Incorrect; Try Again; 3 attempts remaining Part D Where is its center of gravity located? Templates Symbols undo rego Tese keyboard shortouts help m from the end where the 400-N force is applied

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter10: Rotational Motion

Section: Chapter Questions

Problem 77P

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My book has the answers as a) 10N b) 196N But I can't figure out the proper way to work it out. I figured you start by using torque and the weight of the board plus the forces of each support to make a equation for Fn1. Then once you solve for it go back and make a equation for Fn2. But I'm not getting the answers in my book when I do this.
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A beam resting on two pivots has a length of L = 6.00 m and mass M = 89.0 kg.The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 52.5 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. A woman of mass m walking across a beam which is resting on two pivots. The beam is of length L and mass M and the woman is a distance x from the left end of the beam. The first pivot is directly under the left end of the beam and the second pivot is a distance ℓ from the first pivot at a shorter distance than the length of the beam. (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman x meters to the right of the first pivot, which is the origin. (b) Where is the woman when the normal…
A beam resting on two pivots has a length of L = 6.00 m and mass M = 89.0 kg.The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 52.5 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. (a) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. N(b) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.x = m(c) Check the answer to part (e) by computing torques around the first pivot point.x = m(d) Except for possible slight differences due to rounding, is the answer the same?