Question

A displacement B of 100m from the origin at an angle of 37° above the X axis is the result of three successive displacement :b1 of 100 along the negative X axis, b2 of 200m at an angle of 150° above the X axis and a displacement b3. Find b3[ 355m, 353.5]​

Answer 1

Answer:

(353.1 m, -39.8 m)

Explanation:

Let's start by writing the components of the resultant displacement, B:

$B_x = B cos \theta = (100)(cos 37^{\circ})=79.9 m\\B_y = B sin \theta = (100)(sin 37^{\circ})=60.2 m$

We know that this displacement is the result of three successive displacements, which can be written in terms of their components, as follows:

$b_{1x} = -100 m\\b_{1y}=0$

$b_{2x}=(200) cos 150^{\circ} =-173.2 m\\b_{2y} = (200)sin 150^{\circ}=100 m$

and $b_3$, whose components are unknown.

The components of the resultant displacement must be equal to the sum of the components of each vector along each direction, therefore:

$B_x = b_{1x}+b_{2x}+b_{3x}\\\rightarrow b_{3x} = B_x - b_{1x}-b_{2x}=79.9-(-100)-(-173.2)=353.1 m$

and

$B_y= b_{1y}+b_{2y}+b_{3y}\\\rightarrow b_{3y} = B_y- b_{1y}-b_{2y}=60.2-(0)-(100)=-39.8 m$

So, the vector b3 has components (353.1 m, -39.8 m).