The force F acts through the origin. What is the magnitude of F and what angles does it make with x,y,z axes? F = 2.63i + 4.28j-5.92k N

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To determine the magnitude of a force, we'll apply the formula:

|F| = sqrt(a^2 + b^2 + c^2)

where a,b,c are the coefficients of the unit vectors i,j,k.

We'll identify the coefficients: a,b,c.

F = 2.63i + 4.28j-5.92k N

a = 2.63

b = 4.28

c = -5.92

|F| = sqrt[(2.63)^2 + (4.28)^2 + (-5.92)^2]

|F| = 7.75 N

The angle that F makes with x axis is:

cos theta x = a/|F|

cos theta x = 2.63 / 7.75

cos theta x = 70.2 degrees

The angle that F makes with y axis is:

cos theta y = b/|F|

cos theta y = 4.28/7.75

cos theta y = 56.3 degrees

The angle that F makes with z axis is:

cos theta z = c/|F|

cos theta z =  -5.92/7.75

cos theta z = 139.8 degrees

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

The force given here is : F = 2.63i + 4.28j - 5.92k N

Now the magnitude of the force is: sqrt[(2.63)^2 + (4.28)^2 + (-5.92)^2] = 7.75 N

The angle that F makes with x axis is given by:

cos x = 2.63 / 7.75 = .3393

x = arccos (.3393) = 70.16 degrees

The angle that F makes with the y axis is give by:

cos y = 4.28 / 7.75 = .5522

y = arccos (.5522) = 56.47 degrees

The angle that F makes with z axis is given by:

cos z = -5.92 / 7.75 = -0.7638

z = arccos (-0.7638) = 139.80 degrees

Therefore the force has a magnitude of 7.75 and makes angles 70.16, 56.47 and 139.80 degrees with the x, y and z axes respectively.

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