Answer:
I had to search the Figure on Google to solve this question.
a) The magnitude of the force F₃ is:
[tex] F_{3} = 87.47 N [/tex]
And the direction of F₃:
[tex] \alpha = 79.04 ^{\circ} [/tex] Â (with respect to the y-direction, in the third quadrant)
b) P = 4.22 N Â
Explanation:
I had to search the Figure on Google to solve this question.
a) We can find the force of the third person as follows:
[tex] \Sigma F_{x} = F_{1x} + F_{2x} + F_{3x} = 0 [/tex]
[tex] \Sigma F_{y} = F_{1y} - F_{2y} + F_{3y} = 0 [/tex]
So, in x-direction we have:
[tex] \Sigma F_{x} = 45 N*cos(70) + 75 N*cos(20) + F_{3x} = 0 [/tex]
[tex] F_{3x} = -85.87 N [/tex]
In y-direction we have:
[tex] \Sigma F_{y} = 45 N*sin(70) - 75 N*sin(20) + F_{3y} = 0 [/tex]
[tex] F_{3y} = -16.63 N [/tex]
The magnitude of the force F₃ is:
[tex] F_{3} = \sqrt{F_{3x}^{2} + F_{3y}^{2}} = \sqrt{(-85.87 N)^{2} + (-16.63 N)^{2}} = 87.47 N [/tex]
To find the direction of F₃ we need to calculate its angle with respect to the y-direction (in the third quadrant):
[tex] tan(\alpha) = \frac{|F_{3x}|}{|F_{3y}|} = \frac{85.87 N}{16.63 N} [/tex]
[tex] \alpha = 79.04 ^{\circ} [/tex]
b) If the third person exerts the force found in part (a) the car will stop, so the only way for the cart to accelerate at 200 m/s² is that the third person does not exert the force found in a.   Â
To find the weight of the cart​ when it accelerates at 200 m/s², we need to consider: F₃ = 0. Â
First, we need to find the cart's mass. Since the car is moving in the x-direction we have:
[tex] \Sigma F_{x} = F_{1x} + F_{2x} = ma [/tex]
[tex] 45 N*cos(70) + 75 N*cos(20) = m*200 m/s^{2} [/tex]
[tex] m = \frac{45 N*cos(70) + 75 N*cos(20)}{200 m/s^{2}} = 0.43 kg [/tex]
Now, the weight of the cart​ is:
[tex] P = mg = 0.43 kg*9.81 m/s^{2} = 4.22 N [/tex]
I hope it helps you! Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
