In Prob. 3.21, determine the perpendicular distance from point 0 to cable A B.

Name: Problem 27, Chapter 3: Vector Mechanics for Engineers, Statics
Uploaded: 2020-07-22T02:57:15-08:00
Duration: 8 min 1 s
Channel: Kaitlin Coad
Description: In Prob. 3.21, determine the perpendicular distance from point 0 to cable A B.

In Prob. 3.21, determine the perpendicular distance from...

Key Concepts

Vector Projection

Vector projection is used to decompose a vector into components parallel and perpendicular to another vector. In the context of finding the distance from a point to a line, projecting the vector from a point on the line to the external point onto the normal vector of the line isolates the perpendicular component, which represents the shortest distance.

Perpendicular Distance

This concept refers to the shortest distance between a point and a line (or cable, in this case). It is measured along the line that is perpendicular to the given line, ensuring that no component of the distance is parallel to the line.

Point-to-Line Distance Formula

This formula provides a method to compute the distance from a point to a line in a coordinate plane. It typically involves the algebraic representation of the line and uses either coordinate geometry or vector analysis to determine the distance along the perpendicular segment.

Transcript

00:02 All right, so we want to find the perpendicular distance between point a and line b .c.

00:07 To do this, we're going to assume that a force f lies on bc and find the moment of f about point a.

00:14 So the moment is equal to the cross product of the position vector r and the force vector f.

00:21 The magnitude of the moment is equal to the magnitude of the force times the distance.

00:29 Now, f can be positioned anywhere along its line of action b .c.

00:33 To simplify the position vector, we want to place f such that its endpoint lies on point b.

00:41 Now the position vector r is just the vector from a to b.

00:45 This is r.

00:50 It has an x component of 12 feet and y and z components of zero.

01:10 Now the force factor f is the magnitude of f times the unit vector lambda.

01:16 Lambda is equal to the vector bc divided by the magnitude of bc.

01:27 So bc has an x component of negative 12 feet, a y component of 4 .8 feet, and a z component of negative 8 feet.

02:07 The magnitude of bc is the square root of the sum of the components squared.

02:12 So bc is equal to square root of negative 12 feet squared plus 4 .8 feet squared plus negative 8 feet squared, which is equal to 15 .2 feet.

02:43 And now we can find f.

02:50 So f is equal to the vector f is equal to the magnitude of f times the vector b, divided by the magnitude of bc, which is f divided by 15 .2 feet times negative 12 feet times i plus 4 .8 feet j minus 8 feet k.

03:40 This is equal to f times negative 0 .79 i plus 0 .32j minus 0 .53k.

04:03 Alright, now that we have r and f, we can set up the determinant to find the moment...

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