Two parallel 60-N forces are applied to a lever as shown. Dtermine the moment of the couple form

Two parallel 60-N forces are applied to a lever as shown....

Key Concepts

Force Couple

A force couple consists of two parallel forces that are equal in magnitude but opposite in direction and whose lines of action do not coincide. The effect of a couple is to produce a pure rotational effect (a moment) with no net force causing translation, and its magnitude is determined by multiplying the force by the perpendicular distance between the lines of force.

Resolution of Forces

Resolving forces refers to the process of breaking down a force vector into components along mutually perpendicular directions, typically horizontal and vertical. This technique simplifies the analysis of forces and moments by allowing each component’s contribution to be calculated independently, particularly useful when computing the moments about a point.

Moment of a Force

The moment of a force is a measure of its tendency to cause rotation about a pivot or axis. It is calculated as the product of the force magnitude and the perpendicular distance from the pivot to the line of action of the force. This concept is fundamental in understanding rotational equilibrium and the effects of forces on levers and other mechanical systems.

Perpendicular Distance (Lever Arm)

The perpendicular distance, or lever arm, is the shortest distance from the pivot point to the line of action of the force. It plays a critical role in the calculation of the moment of a force, as only the component of the force acting perpendicular to the lever arm contributes to the moment.

Transcript

00:01 3 .71.

00:02 In this problem we have to calculate moment caused by these two forces 40 newton on this rod ac so first of all in first part we have to calculate the moment caused by these two forces by resolving their these forces into their corresponding vertical and horizontal component so first of all let us redraw the segment b c here so that this geometry can be more visualized so let's say this is our segment bc this point is b and this point is c and here the force of 40 newton is acting and it is making an angle of 20 degree with horizontal similarly at point b force of 40 newton is acting in 20 degree angle with horizontal.

01:11 So we can resolve these two components in its corresponding horizontal and vertical component.

01:17 So what will be the value of this 40 newton force in vertical direction will be 40 sine 20 and in horizontal direction it will be 40 cost 20.

01:32 Similarly, this force can be resolved in horizontal component as 40 cost 20 and vertical component as 40 sine 20.

01:45 And we can calculate these values as 40 sine 20 comes out to be 13 .68 and 40 cost 20 comes out to be 37 .58.

02:01 Now this line segment we see has length 2 .6.

02:05 7 tm.

02:08 One thing one can be noted here that line of action of this vertical component and this vertical component will have a perpendicular distance let's say these two line of action intersect at point m so for the vertical component the distance perpendicular distance between these two forces will be bm and that will be from the problem this angle is given to as 55 degrees so this length is 270 cos 55 and this length is 270 sine 55 now calculating the moment caused by the vertical components here one can see that vertical component are causing a anti -clockwise moment and while the horizontal component at point c is in this direction and at point b is in this direction so this is causing a clockwise moment.

03:11 Let us say that clockwise for this problem i'm considering the clockwise moment to be a positive value so anti -clockwise moment is going to be give us negative value so the moment caused by the vertical component will be considered to be negative and moment caused by the horizontal component here will be considered as positive.

03:29 So net moment acting on the element is first of all for the clockwise component that is for horizontal component will be force that is 37 .5a multiplied by perpendicular distance so horizontal forces are separated by a perpendicular distance of 270 sine 55 divided by thousand minus because the vertical components are in anti -clockwise direction minus the magnitude of the force is 40 cost 20 that is 13 .68 multiplied by their perpendicular distance that is 270 cost 55 divided by thousand so this will give the unit of newton meter by calculating these values this comes out to be 6 .20 newton meter since it is a positive quantity so we can say that since our assumption here was clockwise moment to be a positive so this moment final moment on the element will be a clockwise moment this was the part of the problem in part b they are saying that to calculate the same moment by calculating the perpendicular distance between these two forces so again i'm going to redraw the forces here on element bc so this was the element b and this was the force at point b and this was the force at point b and this was the force at point c.

05:23 So for calculating the perpendicular distance one can go through the geometry that is involved in this problem...

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