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Three children are struggling and pulling on a single toy. Two of the children, Abe and Barry, are EACH (individually) pulling with a force of 60 Newtons. The third child, Eric, is pulling with enough force to exactly balance Abe and Barry (no child is winning the tug-o-war for the toy). We can label our forces A, B, and E (from each child's name). These three forces are in equilibrium.

If Abe is pulling toward the North and Barry is pulling toward the East, we want to find the total force R due to Abe and Barry (the Resultant of their forces). Note: in THIS case the two forces and R form a right-triangle (with R as the hypotenuse) - as seen using graphical vector addition. As a convention let's have NORTH upward and EAST rightward on this page.

a. Sketch an ACCURATE free-body diagram (three arrows outward from a point showing LABELED forces A, B, and E - each arrow a reasonable length and correct direction):
b. Sketch a labeled diagram for graphical addition of A and B to get R:
c. Show the calculation of the magnitude of R from the magnitudes of A and B:
d. Give the magnitude of force E (in units of Newtons):
e. Describe the direction of force E in terms of the "cardinal" directions (north, east, south, and/or west):

Sagot :

Solution :

c). [tex]$\vec{F}_A = $[/tex] force applied by Abe

   [tex]$\vec{F}_B = $[/tex] force applied by Barry

   [tex]$\vec{F}_E = $[/tex] force applied by Eric

   [tex]$\vec{F}_R = $[/tex] Resultant force

[tex]$\vec{F}_A $[/tex]  in the vector form can be written as :

         [tex]$\vec{F}_A = 0 \hat{i} + 60 \hat{j}$[/tex]

[tex]$\vec{F}_B $[/tex]  in the vector form can be written as :

         [tex]$\vec{F}_B = 60 \hat{i} + 0 \hat{j}$[/tex]

The resultant,

[tex]$\vec{F}_R= \vec{F}_A+\vec{F}_B $[/tex]

     [tex]$=(0 \hat i + 60 \hat j)+(60 \hat i + 0\hat j)$[/tex]

    [tex]$=60 \hat i + 60 \hat j$[/tex]

[tex]$|\vec{F}_R| = \sqrt{60^2+60^2}$[/tex]

        = 84.853 N

d). As the three forces are in equilibrium, therefore,

[tex]$|\vec F_E| = |\vec F_R|$[/tex]

[tex]$|\vec F_E| =84.853 \ N$[/tex]

e). The direction of the force exerted by Eric is exactly opposite to the direction of the resultant force.

The direction of the resultant force is :

[tex]$\theta = \tan ^{-1}\left(\frac{F_y}{F_x}\right)$[/tex]

   [tex]$ = \tan ^{-1}\left(\frac{60}{60}\right)$[/tex]

  = 45°  north east

The direction of the force E is  45° west or  45° south west.

   

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