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Exercise 8.3 Hints: Getting Started | I'm Stuck Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. Mchild 2 kg (b) Find the normal force acting at the pivot point. N

Exercise 8.3 Hints: Getting Started | I'm Stuck Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. Mchild 2 kg (b) Find the normal force acting at the pivot point. N

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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55 PM Sun Dec 18 Substitute all known quantities. Solve for x substituting the normal force found in part (b). webassign.net à -(75.7 kg)(9.80 m/s2)(2.46 m + x) + (57.8 kg) (9.80 m/s²)(0) - (12.0 kg) (9.80 m/s2)(2.46 m) + n(2.46 m) = 0 -(1.82 x 10³ N-m) - (741.86)x - (289.296 N-m) + (2.46 m)n=0 Exercise 8.3 X Remarks The answers for x in parts (a) and (c) agree except for a small round-off discrepancy. This illustrates how choosing a different axis leads to the same solution. m Hints: Getting Started | I'm Stuck Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. mchild 2 kg (b) Find the normal force acting at the pivot point. N
Transcribed Image Text:55 PM Sun Dec 18 Substitute all known quantities. Solve for x substituting the normal force found in part (b). webassign.net à -(75.7 kg)(9.80 m/s2)(2.46 m + x) + (57.8 kg) (9.80 m/s²)(0) - (12.0 kg) (9.80 m/s2)(2.46 m) + n(2.46 m) = 0 -(1.82 x 10³ N-m) - (741.86)x - (289.296 N-m) + (2.46 m)n=0 Exercise 8.3 X Remarks The answers for x in parts (a) and (c) agree except for a small round-off discrepancy. This illustrates how choosing a different axis leads to the same solution. m Hints: Getting Started | I'm Stuck Suppose a 27.2 kg child sits 1.82 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. mchild 2 kg (b) Find the normal force acting at the pivot point. N
(a) Where should the man sit to balance the seesaw? Apply the second condition of equilibrium to the plank by setting the sum of the torques equal to zero. The first two torques are zero. Let x represent the man's distance from the pivot. The woman is at a distance L/2 from the pivot. Solve this equation for x and evaluate it. (b) Find the normal force n exerted by the pivot on the seesaw. Apply the first condition of equilibrium to the plank, solving the resulting equation for the unknown normal force, n. (c) Repeat part (a), choosing a new axis through the left end of the plank. Compute the torques using this axis, and set their sum equal to zero. Now the pivot and gravity forces on the plank result in nonzero torques. Substitute all known quantities. Solve for x substituting the normal force found in part (b).. pivot + gravity+man+woman-0 0+ 0 Mgx + mg(L/2) - 0 m (L/2) M x 1.878 = (57.8 kg) (2.46 m) 75.7 kg m -Mg-mg - mp9 + n = 0 n = (M + m + mpig = (75.7 kg + 57.8 kg + 12.0 kg) (9.80 m/s²) n= 1425.9 N + pivot gravity + man + woman = 0 -Mg(L/2 + x) + mg(0) - mp19(L/2) + n(L/2) = 0 -(75.7 kg) (9.80 m/s2) (2.46 m + x) + (57.8 kg) (9.80 m/s²) (0) (12.0 kg) (9.80 m/s2) (2.46 m) + n(2.46 m) = 0 -(1.82 x 10³ N-m) - (741.86)x (289.296 N-m) + (2.46 m)n = 0 -
Transcribed Image Text:(a) Where should the man sit to balance the seesaw? Apply the second condition of equilibrium to the plank by setting the sum of the torques equal to zero. The first two torques are zero. Let x represent the man's distance from the pivot. The woman is at a distance L/2 from the pivot. Solve this equation for x and evaluate it. (b) Find the normal force n exerted by the pivot on the seesaw. Apply the first condition of equilibrium to the plank, solving the resulting equation for the unknown normal force, n. (c) Repeat part (a), choosing a new axis through the left end of the plank. Compute the torques using this axis, and set their sum equal to zero. Now the pivot and gravity forces on the plank result in nonzero torques. Substitute all known quantities. Solve for x substituting the normal force found in part (b).. pivot + gravity+man+woman-0 0+ 0 Mgx + mg(L/2) - 0 m (L/2) M x 1.878 = (57.8 kg) (2.46 m) 75.7 kg m -Mg-mg - mp9 + n = 0 n = (M + m + mpig = (75.7 kg + 57.8 kg + 12.0 kg) (9.80 m/s²) n= 1425.9 N + pivot gravity + man + woman = 0 -Mg(L/2 + x) + mg(0) - mp19(L/2) + n(L/2) = 0 -(75.7 kg) (9.80 m/s2) (2.46 m + x) + (57.8 kg) (9.80 m/s²) (0) (12.0 kg) (9.80 m/s2) (2.46 m) + n(2.46 m) = 0 -(1.82 x 10³ N-m) - (741.86)x (289.296 N-m) + (2.46 m)n = 0 -
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