Given the following: F1 = 250 kN A = 60 deg F2 = 100 kN B = 70 deg F3 = 300 kN C

Question

Given the following:

F1 = 250 kN     A = 60 deg        

F2 = 100 kN     B = 70 deg     

F3 = 300 kN     C = 20 deg        

Using the Figure below, determine the magnitude, in kN, the direction, in degrees (as a positive number less than 90 degrees relative to the x-axis), and quadrant of the resultant for the coplanar concurrent force system shown below. Input your answers in the order shown by filling in the associated blanks with your answer. Express the value of the Quadrant as a whole number between 1 and 4. Express the resultant and angle to one decimal place. Example: 17.1 DO NOT INCLUDE UNITS WITH YOUR ANSWER  Note: When rounding your answers, round as the last step in the calculation process.

Resultant = ___ kN

Angle or direction = ___degrees

Quadrant, ______

Explanation / Answer

F1 = 250 kN, A = 60 deg

F1 = 250*cos 60 deg i + 250*sin 60 deg j

F2 = 100 kN, B = 70 deg

F2 = -100*cos 70 deg i - 100*sin 70 deg j

F3 = 300 kN, C = 20 deg

F3 = 300*cos 20 deg i - 300*sin 20 deg j

So Net force will be:

Fnet = (250*cos 60 deg i + 250*sin 60 deg j) + (-100*cos 70 deg i - 100*sin 70 deg j) + (300*cos 20 deg i - 300*sin 20 deg j)

Fnet = (125 - 34.20 + 281.90) i + (216.51 - 93.97 - 102.61) j

Fnet = 372.7 i + 19.93 j

|Fnet| = sqrt (372.7^2 + 19.93^2)

|Fnet| = 373.23 N

Direction = arctan (19.93/372.7)

Angle of direction = 3.06 deg

Since both (Fnet)x and (Fnet)y are positive, which means net force is in first quadrant.

Let me know if you have any doubt.

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