In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force F A of magnitude 222 N, and Charles pulls with force F c of magnitude 188 N. Note that the direction of F c is not given. What is the magnitude of Betty's force F B if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium? Alex Charles Betty 140°

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In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The
tire remains stationary in spite of the three pulls. Alex pulls with force É A of magnitude 222 N, and Charles pulls with force
magnitude 188 N. Note that the direction of F c is not given. What is the magnitude of Betty's force F B if Charles pulls in (a) the
c of
direction drawn in the picture or (b) the other possible direction for equilibrium?
Alex
Charles
140°
Betty
(a) Number
i
Units
(b) Number
i
Units
Transcribed Image Text:In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force É A of magnitude 222 N, and Charles pulls with force magnitude 188 N. Note that the direction of F c is not given. What is the magnitude of Betty's force F B if Charles pulls in (a) the c of direction drawn in the picture or (b) the other possible direction for equilibrium? Alex Charles 140° Betty (a) Number i Units (b) Number i Units
Expert Solution
Step 1

Given.FA=222NFC=188NFrom figure 180°-140°=40°then 90°-40°=50°

Advanced Physics homework question answer, step 1, image 1

(a) Net force in x direction, Fx=0-FAcos 50°+FCcosθ=0FCcosθ=FAcos 50°cosθ=FAcos 50°FC=222 cos50°188=0.759θ=cos-1(0.759)=40.6°Net force in y direction, Fy=0FA sin50°-FB+FCsinθ=0FB=FA sin50°+FCsinθFB=222sin50°+188sin(40.6°)FB=292.41 N

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