This question is solved using an available similar problem as data provided for the forces was not given.
Repeat the same steps outlined for your problem.
Regards.
Answer:
F = 1.598 KN , Q = 90 degree (+ y-axis)
Explanation:
Sum of Forces in x-direction to the left (+)
2 cos (30) + 3cos (60) + F*cos (Q) = F_a ..... 1
Sum of Forces in y-direction to the up (+)
2 sin (30) + F*sin (Q) - 3 sin (60) ...... 2
Using Eq 2 and solve:
F*sin (Q) = 1.598 KN
F_min when sin (Q) is max, max possible value of sin(Q) = 1 @ Q = 90 degrees.
Hence,
F_min = 1.598 KN
Using Eq 1 @ Q = 90 degrees and F = 1.598 KN:
F_a = 2 cos (30) + 3cos (60) = 3.2 KN
<h3>WATER</h3>
Explanation:
<h2>#CARRYINGTOLEARN:)</h2>
Answer:
Explanation:
Given
winning car accelerates with a and its final velocity is v
considering they both start from rest
time taken by winning car is
v=u+at
where u=initial velocity
a=acceleration
t=time
Now loosing car is accelerating with 
Distance traveled by loosing car in time t
Thus distance d traveled by loosing car is given by
