In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on a

Question

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 Upper F Overscript right-arrow EndScripts Subscript Upper A of magnitude 204 N, and Charles pulls with force Upper F Overscript right-arrow EndScripts Subscript Upper C of magnitude 186 N. Note that the direction of Upper F Overscript right-arrow EndScripts Subscript Upper C is not given. What is the magnitude of Betty's force Upper F Overscript right-arrow EndScripts Subscript Upper B if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium?

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 FA of magnitude 204 N, and Charles pulls with force Fc of magnitude 186 N. Note that the direction of Fc is not given. What is the magnitude of Betty's force FB if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium?

Explanation / Answer

let the direction along which betty is pully the tire is negative y axis

then positive x axis will be perpendicular to the direction in which betty is pulling.

now , assume that i and j are unit vectors along +ve x axis and +ve y axis repsectively

let magnitude of betty's force=F N

then in vector notation, betty's force=Fb=-F j

angle made by the direction of force applied by Alex , with -ve x axis=139-90=49 degrees

then in vector notation, force applied by alex=204*(-cos(49) i + sin(49) j)

=-133.836 i +153.96 j N

let angle made by charles with +ve x axis is theta.

then force applied by charles in vector notation=186*(cos(theta) i + sin(theta) j)

as the tire is in equilibrium, force along x axis and y axis are both zero

hence equating the components:

equating x component of total force with zero:

-133.836+186*cos(theta)=0

==>cos(theta)=0.71954

==>theta=43.982 degrees

equating y component of total force with zero,

-F + 153.96+186*sin(theta)=0

==>F=153.96+186*sin(43.982)

==>F=283.124 N

hence force applied by betty is 283.124 N

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