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What is the magnitude and direction of the resultant force?A force of F1 of 36 N pulls...

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takenotes | Student, Grade 9 | (Level 1) Honors

Posted July 9, 2012 at 4:15 AM via web

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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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najm1947 | Elementary School Teacher | (Level 1) Valedictorian

Posted July 9, 2012 at 10:19 AM (Answer #1)

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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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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 24, 2012 at 2:23 PM (Answer #2)

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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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