Given the force F = 3.5i + 3.5j - 7k N determine the magnitude of the force and the angles with axis . F passes through origin .

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We have the force F defined as F = 3.5i + 3.5j - 7k N

Now we need to find the magnitude and the angle that the vector F makes with the axes.

For a vector F= ai + bj + ck

The magnitude |F| is given by sqrt(a^2 + b^2 + c^2)

=> sqrt [(3.5)^2 + (3.5)^2 + (-7)^2]

=> 8.57 N

|F| = 8.57 N

The angle that F makes with x axis is given by a/ |F|

=> cos x = 3.5/ 8.57

=> cos x = 0.4084

=> x = arccos (0.4084) = 65.89 degrees

The angle that F makes with the y axis is given by b/|F|

=> cos y = 3.5/ 8.57

=> cos y = 0.4084

=> y = arccos (0.4084)

=> y = 65.89 degrees

The angle that F makes with the z axis is given by c/|F|

=> cos z =  -7/8.57

=> cos z = -0.8168

=> z = arccos ( -0.8168) = 35.23 degrees

Therefore the magnitude of F is 8.57 N and teh angles made with the x, y and z axes is 65.89, 65.89,  and 35.23 degrees respectively

Top Answer

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 write F and we'll identify the coefficients: a,b,c.

F = 3.5i + 3.5j - 7k N

a = 3.5

b = 3.5

c = -7

|F| = sqrt[(3.5)^2 + (3.5)^2 + (-7)^2]

|F| = 8.6 N

The angle that F makes with x axis is:

cos theta x = a/|F|

cos theta x = 3.5/ 8.6

cos theta x = 0.406

The angle that F makes with y axis is:

cos theta y = b/|F|

cos theta y = 3.5/ 8.6

cos theta y = 0.406

The angle that F makes with z axis is:

cos theta z = c/|F|

cos theta z =  -7/8.6

cos theta z = -0.81

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The given force  F = 3.5i+3.5j-7k  N.

The magnitude of the force F = xi+yj+zk is sqrt(x^2+y^2+z^2).

Therefore the magnitude of 3.5i+3.5j-7k N is sqrt{3.5^2+3.5^2 +(-7)^2)} = 8.5732N

And the direction of the force with x axis ( in the direction of unit vector i)  is  given by cosx =3.5/8.5732 = 0.4082.

Or  x =  arc cos (0.4082) = 65.9052 deg.

Or x = 1.15026 radians.

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