Answer:
R = 3124.3 P
α = -10.38°
Explanation:
Look at the attached graphic
We take the forces acting on the wagon in the x-y plane
x axis : x axis: parallel to the road
y axis :perpendicular to the road
F₁ = 125 P at an angle of 29° with respect to the x axis
F₂ = 3030 P , at an angle of −12° with respect to the x axis
x-y components F₁ (donkey rope force connected to the wagon)
F₁x= 125*cos(29)° =109.33 P
F₁y= 125*sin(29)° =66.6 P
x-y components F₂(dog rope force connected to the wagon)
F₂x= 3030*cos(-12)° =2963.78 P
F₂y= 3030*sin(-12)° = - 629.97 P
Calculation of the components of resultant force (R)
Rx= F₁x +F₂x
Rx=109.33 P +2963.78 P
Rx= 3073.11
Ry=F₁y + F₂y
Ry= 66.6 P - 629.97 P
Ry= - 563.37 P
Magnitude of the resultant force (R)
[tex]R= \sqrt{(R_{x})^{2} + (R_{y})^{2} }[/tex]
[tex]R= \sqrt{(3073.11)^{2} + (-563.37)^{2} }[/tex]
R = 3124.3 P
Direction of resultant force (α)
[tex]\alpha =tan^{-1} (\frac{R_{y} }{R_{x} } )[/tex]
[tex]\alpha =tan^{-1} (\frac{-563.37 }{3073.11 } )[/tex]
α = -10.38°
