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

(353.1 m, -39.8 m)

Step-by-step 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).