The handpiece of a miniature industrial grinder weighs 2.4 N...

The handpiece of a miniature industrial grinder weighs 2.4 N , and is center of grivity is located on the $y$ axis. The head of the handpiece is offset in the $x=$ plane in such a way that line $B C$ forms an angle of $25^{\circ}$ with the $x$ direction. Show that the weight of the handpiece and the two couples $\mathbf{M}_1$ and $\mathbf{M}_2$ can be replaced with a single equivalent force. Further assuming that $M_1=0.068$ N -mand $\mathrm{M}_2=0.065 \mathrm{~N} \cdot \mathrm{~m}$, determine ( $a$ ) the magnitude and the direction of the equivalent force, $(b)$ the point where its line of action intersects the $x z$ plane. (FIGURE CAN'T COPY)

Key Concepts

Line of Action

The line of action of a force is an imaginary line that extends indefinitely in both directions along the direction of the force, passing through its point of application. In statics, knowing the line of action is critical because if the net moment of a force about any point on its line of action is zero, the force can be 'moved' along this line without affecting the equilibrium of the system. This concept is fundamental when determining the point of application of an equivalent force that represents a distributed load or a force-couple system.

Force-Couple System

A force-couple system consists of a force acting at a point along with a couple (or moment), which is a pair of equal and opposite forces whose lines of action do not coincide. The system can be analyzed by separating the effects of the force and the couple, and under certain conditions, can be replaced by a single equivalent force that produces the same net external effects on the body. This concept is key in mechanics when simplifying the net action of multiple forces and moments acting on a rigid body.

Equivalent Force

The equivalent force is the single resultant force that can be substituted for a more complicated system of forces and couples in order to simplify the analysis of the system. This is achieved by ensuring that both the net force and the net moment about an appropriate reference point are the same as those produced by the original system. The idea is to relocate the effect of the couple into a single force whose line of action is chosen to produce no additional moment, thus preserving static equilibrium.

Moment

A moment, often referred to as torque, is a measure of the tendency of a force to rotate an object about an axis, a pivot, or a point. It is calculated as the cross product of the position vector (from a chosen reference point to the point of application of the force) and the force vector itself. Moments are essential in determining how forces cause rotational effects in static equilibrium problems and in reducing complex force systems to simpler ones.

Transcript

00:01 Okay, here what we are asked to do is we are asked to find the x and y components of the center of gravity of this tool.

00:08 Now, it's broken into two pieces, so i'm going to label the center of gravity and the coordinates in different colors.

00:17 Now, for this leftmost piece, it's just going to be in the dead solid middle.

00:21 And that means the x and y components are going to be half of both of these pieces.

00:27 So x1 is going to be two centimeters because it's half of the four and y1 is going to be 5 .5 centimeters because it's half of the 11.

00:37 Now normally we'd want to want to convert to si, but we do get unit cancellations here.

00:44 So we'll just leave this here.

00:46 Now, same thing is going to be true for the other piece.

00:50 It's x is just going to be half of the 16.

00:53 So x2 is going to be eight centimeters.

00:57 And this, i forgot to label this, but this was three centimeters.

01:01 Now, the y2, you're going to have to tack on up to there, which will be eight centimeters.

01:09 So you're going to have to do eight centimeters plus half of the three centimeters.

01:14 So y2 is going to be equal to eight plus half three, which is one point five.

01:19 So nine point five centimeters.

01:21 Now, from there, we can just do x1 equals, or x center of mass equals x1 plus x2, m2 over m1 plus m2.

01:36 And so that's going to be two times 40, because that's the first piece, plus 8 times 80, because that's the second piece, all over 120.

01:50 And this is going to be 80 plus 640 over 120, which is going to be 720 over 120.

02:04 And so my x center of mass is going to be 720 divided by 120, which is just 6 centimeters.

02:13 Okay.

02:14 And then my y center mass, similar sort of thing, y1m1 plus y1.

02:20 I2, m2, and it's all over m1 plus m2.

02:25 And so, so then we're going to have, so the y was 5 .5 times 40.

02:36 And the y2 was 9 .5 times 80.

02:43 And my total mass is 120.

02:45 So i'll just throw that into the calculator really quick, and that's going to be 5 .5...

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