Question #daa19

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Question #daa19

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Answer 1 · 2017-06-07T07:33:12

$Please refer to the description section and check the operations.$ $"Please refer to the description section and check the operations."$

Explanation:

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The forces and their directions are described in figure-1.

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We can solve the problem in many ways.
In this solution we used the method of dividing the forces into horizontal and vertical components.

$The components of F_{3} are :$ $"The components of " F_3 " are :"$
$F_{3x} = F_{3} \cdot cos (30) = 50 \sqrt{3} . \frac{\sqrt{3}}{2} = \frac{50.3}{2} = \frac{150}{2} = 75 N$ $F_("3x")=F_3*cos(30)=50sqrt (3). sqrt(3)/2=(50.3)/2=150/2=75N$

$F_{3y} = F_{3} . sin (30) = 50 \sqrt{3} \cdot \frac{1}{2} = 25 \sqrt{3} N$ $F_("3y")=F_3.sin(30)=50sqrt(3)*1/2=25sqrt(3)N$

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$The components of F_{1} are :$ $"The components of " F_1 " are :"$

$F_{1x} = F_{1} \cdot cos (0) = 100.1 = 100 N$ $F_("1x")=F_1*cos(0)=100.1=100N$

$F_{1y} = F_{1} . sin (0) = 100.0 = 0 N$ $F_("1y")=F_1.sin(0)=100.0=0N$

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$The components of F_{2} are :$ $"The components of " F_2 " are :"$
$F_{2x} = F_{2} \cdot cos (30) = 100 \sqrt{3} . \frac{\sqrt{3}}{2} = \frac{100.3}{2} = \frac{300}{2} = 150 N$ $F_("2x")=F_2*cos(30)=100sqrt (3). sqrt(3)/2=(100.3)/2=300/2=150N$

$F_{2y} = - F_{2} . sin (30) = 100 \sqrt{3} \cdot \frac{1}{2} = - 50 \sqrt{3} N$ $F_("2y")=-F_2.sin(30)=100sqrt(3)*1/2=-50sqrt(3)N$

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Figure-5 shows the vector sum of the horizontal components of the forces.

$Sigma F_("x")=F_("1x")+F_("2x")+F_("3x")$
$Sigma F_("x")=100+150+75=325N$

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Figure-6 shows the vector sum of the vertical components of the forces.

$Sigma F_("y")=F_("1y")+F_("2y")+F_("3y")$
$Sigma F_("y")=0-50sqrt(3)+25sqrt(3)=-25sqrt(3)N$

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Figure 7 shows the resultant vector and direction.

$F_r=sqrt((Sigma F_x)^2+(Sigma F_y)^2)$

$F_r=sqrt((325)^2+(-sqrt(25))^2)$

$F_r=sqrt(105625+625)$

$F_r=325.96N$

We must specify the angle of the resultant vector.

$tan theta=(Sigma F_y) / (Sigma F_x)$

$tan theta=-(25sqrt(3))/325=-(sqrt(3))/13=-0.13323468$

$theta=-7.59^o$

Answer 2 · 2017-06-07T07:43:28

The resultant force is $=327.9N$ in the direction $=7.6º$ of the center rope

Explanation:

$cos30=sqrt3/2$

$sin30=1/2$

Resolving in thedirection parallel to the $100N$ force

$F_h=100+50sqrt3cos30 +100sqrt3cos30$

$=100+50sqrt3*sqrt3/2+100sqrt3*sqrt3/2$

$=100+75+150=325N$

Resolving perpendicular to the $100N$ force

$F_p=100sqrt3sin30-50sqrt30sin30$

$=100sqrt3*1/2-50sqrt3*1/2$

$=50sqrt3-25sqrt3$

$=25sqrt3$

The resultant force is

$F=sqrt(F_h^2+F_p^2)$

$=sqrt(325^2+(25sqrt3)^2)$

$=sqrt107500$

$=327.9N$

The direction is

$theta=arctan((25sqrt3)/325)$

$=arctan(sqrt3/13)$

$=7.6º$

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