What is the magnitude and direction of the resultant force? A force of F1 of 36 N pulls at an angle of 20° above due east. Pulling in the opposite direction is a force F2 of 48 N acting at an angle of 42° below due west. I have no idea on how to set up this problem. Any help is greatly appreciated.
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The force F1 component to East = 36cos20 = 33.83N
The force F1 component to North = 36sin20 = 12.31
The force F2 component to West = 48cos42 = 35.67N
The force F2 component to South = 48sin42 = 32.12N
Let take forces to East and north as positive.
Then forces to West and South are negative forces to East and North forces.
Total force to East = 33.83-35.67 = -1.84N
Total force to North = 12.31-32.12 = -19.81N
Resultant force = sqrt[(-1.84)^2+(-19.81)^2] = 19.89N
Direction of resultant force = tan^(-1) (-19.81/-1.84)
= 84.69 deg.
The resultant force is a force of 19.89N to the direction of 84.69 deg. below due west.
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ANYONE HELP ME WITH THIS QUESTION
F1 F2 F3 F4 θ1 θ2 d1 d2 d3 21 35 10 52 115° 260° 1.5 2.4 4.7
i. Calculate the magnitude and direction of the resultant force.
ii.Find the perpendicular distance of the line of action for the resultant force.
iii. Give the magnitude, direction and line of action for the equilibrant.
An object is subject to four forces as shown in the diagram. The object rotates about point O.
Let the due east force be positive and due west force be negative, similarly let upwards force be positive and downwards force be negative then the components of the two forces in east and upwards directions will be as under:
Net eastward force = 36cos(20) - 48cos(42) = -1.84 N, it is a westwards force as it is negative
Net upward force = 36sin(20) - 48sin(42) = -19.81 N, it is a downward force as it is negative
The magnitude of the resultant force = sqrt{(-1.84)^2+(-19.81)^2} = 19.9 N
Direction of resultant force = tan-1{(-18.91)/(-1.84)} = 84.7 degree below due west
Resultant force is 19.9 N working 84.7 degree below due west
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