The magnitude of the force vector [latex]F_B[/latex] is 2 kN. Express it in terms of scalar components.

The strategy is to determine the unit vector collinear with [latex]F_B[/latex] and then express the force in terms of this unit vector. The radius vector collinear with [latex]F_B[/latex] is [latex]r _{B D}=(4-5) i +(3-0) j +(1-3) k \text { or } r _{B D}=-1 i +3 j -2 k[/latex]. The magnitude is [latex]\left| r _{B D}\right|=\sqrt{1^2+3^2+2^2}=3.74[/latex]. The unit vector is [latex]e _{B D}=\frac{ r _{B D}}{\left| r _{B D}\right|}=-0.2673 i +0.8018 j -0.5345 k[/latex] The force is [latex]\begin{aligned}F _B & =\left| F _B\right| e _{B D}=2 e _{B D}( kN ) F _B=-0.5345 i +1.6036 j -1.0693 k \\& =-0.53 i +1.60 j -1.07 k~( kN )\end{aligned}[/latex]

Question 2.161: The magnitude of the force vector FB is 2 kN. Express it in ......

Engineering Mechanics Statics – Solution Manual [1686042]

The magnitude of the force vector $F_B$ is 2 kN. Express it in terms of scalar components.

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The magnitude of the force vector FB is 2 kN.

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Step 1:

To determine the unit vector collinear with the force vector FB, we first need to find the radius vector BD. This vector represents the displacement from point B to point D. We can calculate BD by subtracting the coordinates of point B from the coordinates of point D, resulting in a vector of (-1, 3, -2).

Step 2:

Next, we calculate the magnitude of the radius vector BD, which is the length of the vector. Using the formula for vector magnitude, we find that the magnitude of BD is equal to the square root of the sum of the squares of its components, which in this case is sqrt(1^2 + 3^2 + 2^2) = 3.74.

Step 3:

To obtain the unit vector eBD, we divide the radius vector BD by its magnitude. This will give us a vector that has the same direction as BD but a magnitude of 1. Dividing each component of BD by its magnitude, we find that eBD is equal to (-0.2673, 0.8018, -0.5345).

Step 4:

Finally, to express the force FB in terms of the unit vector eBD, we multiply the magnitude of FB by eBD. This gives us a vector that has the same direction as FB but a magnitude proportional to the magnitude of FB. Multiplying each component of eBD by the magnitude of FB, we find that FB is equal to (-0.5345, 1.6036, -1.0693) kN.

In summary, the strategy used in this problem was to find the unit vector collinear with FB by determining the radius vector BD, calculating its magnitude, and dividing it by the magnitude to obtain the unit vector. Finally, we expressed the force FB in terms of the unit vector by multiplying its magnitude by the unit vector.