Answer: Force of magnitude 20.22lbf to the right at angle 8.5°
Explanation: The image named "Untitled" below shows the forces acting on the cart.
The second image, "Untitled2", demonstrate the vector diagram of the forces acting on the kart, in which:
f is force forward
f(left) is pull to the left
f(R) is the resulting force necessary for the kart to move
The drag force is not draw on the vector diagram because its value is already part of the force forward.
As we can see, the diagram forms a right triangle with f(R) as hypotenuse.
Using Pytagorean Theorem:
[tex][f(R)]^{2}=f^{2}+[f(left)]^{2}[/tex]
[tex][f(R)]^{2}=20^{2}+3^{2}[/tex]
[tex][f(R)]^{2}=409[/tex]
[tex]f(R)=\sqrt{409}[/tex]
f(R) = 20.22
Force forward equals 20.22lbf to the right.
The angle formed form the forward direction and the resulting force is
[tex]cos\theta=\frac{20}{20.22}[/tex]
cosθ = 0.989
θ = cos⁻¹(0.989)
θ = 8.5°
The forward force is to the right at angle 8.5° and magnitude 20.22lbf.
