The towline exerts a force of p = 4 kn at the end of the 20-m-long crane boom. if u = 30, determine the placement x of the hook
1 answer:
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Answer:
a. Fnet =37.67N
b. The direction = 133.4 from the x axis counter clockwise.
c. Option 2
Explanation:
Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.
Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..
Given that F1 is 71N at 20°, then it is in the first quadrant.
a. Fnet= F1+F2+F3
Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j
Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j
Resolving the vectors into x and y components.
Fnet= -2.04i+37.62j
Magnitude of the vector
Fnet= √((-2.04)^2+(37.62)^2)
Fnet= 37.67N
Fnet is approximately 38N.
b. Direction of the Fnet.
Angle=arctan(y/x)
Angle=arctan(-37.61/2.04)
Angle= -43.37°
The angle is in the negative x axis and positive y axis.
Then the direction becomes 180-43.37
Therefore, the direction of the net force is 133.37°.
c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.
Fnet=ma
Fnet= mv/t
So the velocity is in the direction of the Fnet.
If the current takes him downstream we must find the resultant vector of the velocities: 
Then if the river is 3000 m-wide the swimmer will have to pass:
1.3520747 · 300 = 4056.14 m t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
1.3520747 · 300 = 4056.14 m t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.