Answer to Question #240042 in Molecular Physics | Thermodynamics for Sakura

Question #240042

Four sled dogs are pulling a 1000kg load. Sled dog A is pulling the load at 20N, 10 degrees north of east. Sled dog B is pulling the load at 55N, 70 degrees north of east. Sled dog C is pulling the load at 45N, 33 degrees north of west, while sled dog D exerts 30N at a direction of 80 degrees south of west. What is the resultant force acting on the load?


1
Expert's answer
2021-09-21T14:18:14-0400

Let's first find the xx- and yy-components of the resultant force:

Fres,x=20 Ncos10+55 Ncos70+45 Ncos(18033)+30 Ncos(180+80)=4.44 N,F_{res,x}=20\ N\cdot cos10^{\circ}+55\ N\cdot cos70^{\circ}+45\ N\cdot cos(180^{\circ}-33^{\circ})+30\ N\cdot cos(180^{\circ}+80^{\circ})=-4.44\ N,

Fres,y=20 Nsin10+55 Nsin70+45 Nsin(18033)+30 Nsin(180+80)=50.12 N.F_{res,y}=20\ N\cdot sin10^{\circ}+55\ N\cdot sin70^{\circ}+45\ N\cdot sin(180^{\circ}-33^{\circ})+30\ N\cdot sin(180^{\circ}+80^{\circ})=50.12\ N.

We can find the magnitude of the resultant force from the Pythagorean theorem:


Fres=Fres,x2+Fres,y2=(4.44 N)2+(50.12 N)2=50.32 N.F_{res}=\sqrt{F_{res,x}^2+F_{res,y}^2}=\sqrt{(-4.44\ N)^2+(50.12\ N)^2}=50.32\ N.

We can find the direction of the resultant force from the geometry:


θ=tan1(Fres,yFres,x),\theta=tan^{-1}(\dfrac{F_{res,y}}{F_{res,x}}),θ=tan1(50.12 N4.44 N)=85 N of W.\theta=tan^{-1}(\dfrac{50.12\ N}{-4.44\ N})=85^{\circ}\ N\ of\ W.

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