Question

A pole, 6 m long, lies along the x-axis with one end at the origin. A rope is attached to the other end (point A), which runs to a point on the wall (point B), given by the coordinates (0,-1,2) in meters. Also at point A, there is a force given by F = 400 i - 200 j + 500 k (in N). Determine the projected component of this force (in N) acting along the rope AB.

Answer 1

The value is
`|P| =  187.4 \ N`

Step-by-step explanation:

From the question we are told that

The length of the first rope is L = 6 m

The first rope lie on the x-axis

The end point of the first rope is A

Now the vector of A will be
`\vec {OA}=  6i`

The point of the second rope on the wall is B

The coordinate for the point of the second rope on the wall is (0,-1,2)

The vector of B will be
`\vec {OB} =  -j+2k`

Now the coordinate of rope AB is mathematically represented as

`\vec {AB} =  \vec{OB} -\vec{OA}`

=>
`\vec {AB} =  -j+2k -6i`

=>
`\vec {AB} =  -6i -j+2k`

Generally the magnitude of the rope AB is mathematically evaluated as

`|\vec{AB}| = √((-6^2) +(-1)^2 + (2)^2)`

`|\vec{AB}| = √(41) \ m`

Generally the unit vector rope AB is mathematically evaluated as

`\vec r = \frac{\vec {AB}}{|\vec{AB}|}`

=>
`\vec r =  (1)/(√(41) ) * [-6i -j+ 2k]`

From the question we are told that there is a force acting at point A and the force is

`F = 400 i - 200 j + 500 k`

Generally the projected component of this force (in N) acting along the rope AB is mathematically represented as

`P =  \vec F \cdot  \vec r`

=>
`P  = 400 i - 200 j + 500 k  \ * \  (1)/(√(41) ) * [-6i -j+ 2k]`

Note ( i . i = 1 ) , (j . j = 1) , (k . k = 1)

So
`P =  (1)/(√(41) ) [-400 * 6 + 200 * 1 + 500 *2]`

=>
`P =  -187.4 \ N`

So the magnitude of the projected component of this force (in N) acting along the rope AB is

`|P| =  187.4 \ N`