Question

(b) In the adjacent figure, find the value of F and theta so that the sum
of the vectors will be zero​

Answer 1

Explanation:

see the attachment. ......

Answer 2

Answer:

The value of Force(F) is 50N and the value of theta is 57•

Explanation:

Given picture is a representation of 3 Force vectors in a two dimensional co-ordinate system.To solve the given question,we are required to resolve the vectors in two 2 components along both the axes.

Resolution of Unknown force along x-axis is

$F_{x}=Fcos\theta$

Similarly resolution of 2nd force vector along x-axis is

$(F_{2})_{x}=20\sqrt{2}sin45° \\ = 20 \sqrt{2} \times \frac{1}{ \sqrt{2} } = 20N$

The third force acts along the negative x direction with a magnitude of 10N

Even the component of Second force acts along the negative x direction.For equilibrium we have an relation,

$ΣF_{x}=0$

Therefore,the resultant of all the forces along the horizontal is zero.

$=>Fcos\theta-20-10=0 \\ = >Fcos\theta = 30 N$

Simlarly the resolution of first and second forces along y-axis are written as follows.

$F_{y}=Fsin\theta \\ </strong></p><p><strong>[tex]F_{y}=Fsin\theta \\ (F_{2})_{y}=20 \sqrt{2} \sin(45) = 20N$

A force of magnitude 60N acts along the negative y direction hence resultant of the forces along the y axis becomes zero for equilibrium.

$=>Fsin\theta+20-60=0</p><p> \\ = >Fsin\theta = 40 N$

Now dividing both the horizontal and verical components of the unknown force we get

$= > \frac{Fsin\theta}{Fcos\theta}=\frac{40}{30} \\ = > \tan( \theta ) = \frac{4}{3} \\ = > \theta = 57 {}^{0}$

If this is known then,

$\sin( \theta ) = \frac{4}{ \sqrt{ {3}^{2} + {4}^{2} } } = \frac{4}{5}$

Substituting this value in the vertical component of unknown force we get

$F \times \frac{4}{5} = 40 N \\ = > F = 50N$

Therefore,the value of Force(F) is 50N and the value of theta is 57•

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