) If FB = 560 N and FC = 700 N , determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

Magnitude of FR = 1174.56°N

α=67

β=92

γ=157

Explanation:

From the diagram, we know that

A:(0i+0j+6k)

B:(2i-3j+0k)

C:(3i+2j+0k)

First, we'll find the position vectors from A to B and A to C.

rAB ={(2−0)i+(−3−0)j+(0−6)k}

rAB ={2i−3j−6k} m

rAC ={(3−0)i+(2−0)j+(0−6)k}

rAC ={3i+2j−6k} m

From this point, we'll then find the magnitude of each position vector so that we can use it to find the unit vector in the next step.

Magnitude of rAB = √(2)² + (-3)² + (-6)²

Magnitude of rAB = √(4 + 9 + 36)

Magnitude of rAB = √49

Magnitude of rAB =7m

Magnitude of rAC = √(3)² + (2)² + (-6)²

Magnitude of rAC = √(9 + 4 + 36)

Magnitude of rAC = √49

Magnitude of rAC =7m

Now, we'll find the unit vector for each force. Remember, this is found by dividing each component of the position vector by it’s magnitude.

uAB = (2i/7 -3j/7 - 6k/7)

uAC = (3i/7 + 2j/7 - 6k/7)

FB = 560 * uAB

FB = 560 * (2i/7 -3j/7 - 6k/7)

FB = {160i - 240j - 480k}N

FC = 700 * uAC

FC = 700 * (3i/7 + 2j/7 - 6k/7)

FC = {300i + 200j - 600k}N

We'll add both forces together to find the resultant force

FR = FC + FB

FR = {460i - 40j - 1080k}N

Then, we'll find the magnitude of the resultant force to get the coordination direction angle.

Magnitude of FR = √(460)² + (-40)² + (-1080)²

Magnitude of FR = 1174.56N

Finding the coordination angles

α=cos-1(460/1174.56) = 67° --- Approximated

β= cos-1(-40/1174.56) = 92° ---- Approximated

γ= cos-1(-1080/1174.56) = 157° ---- Approximated

answer; /// the above statement is (true);

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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