Question

A donkey and a large dog are both hooked to tow-ropes connected to a wagon stuck in the mud in the middle of a country road. The donkey pulls with a force of 125 pounds at an angle of 29∘ with respect to the road. The dog pulls with a force of 3030 pounds at an angle of −12∘ with respect to the road. Find the magnitude and direction of the resultant force. Include appropriate units with your answers.

Answer 1

R = 3124.3 P

α = -10.38°

Step-by-step 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)

`R= \sqrt{(R_(x))^(2) + (R_(y))^(2) }`

`R= \sqrt{(3073.11)^(2) + (-563.37)^(2) }`

R = 3124.3 P

Direction of resultant force (α)

`\alpha =tan^(-1) ((R_(y) )/(R_(x) ) )`

`\alpha =tan^(-1) ((-563.37 )/(3073.11 ) )`

α = -10.38°