Question

Question #daa19

Answer 1

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

Explanation:

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

• 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 :"`
`F_("3x")=F_3*cos(30)=50sqrt (3). sqrt(3)/2=(50.3)/2=150/2=75N`

`F_("3y")=F_3.sin(30)=50sqrt(3)*1/2=25sqrt(3)N`

`"The components of " F_1 " are :"`

`F_("1x")=F_1*cos(0)=100.1=100N`

`F_("1y")=F_1.sin(0)=100.0=0N`

`"The components of " F_2 " are :"`
`F_("2x")=F_2*cos(30)=100sqrt (3). sqrt(3)/2=(100.3)/2=300/2=150N`

`F_("2y")=-F_2.sin(30)=100sqrt(3)*1/2=-50sqrt(3)N`

• 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`

• 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`

• 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

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`