While a sagging porch is leveled and repaired, a screw jock is asd to support the front of the p

While a sagging porch is leveled and repaired, a screw jock...

Key Concepts

Principle of Transmissibility

The principle of transmissibility states that the external effect of a force on a rigid body is unaffected if the force is moved along its line of action. This property is critical in statics, as it allows forces to be repositioned in analyses without altering the overall equilibrium provided the corresponding moments are properly adjusted.

Force-Couple System

A force-couple system is a method in statics for representing the effect of loads on a rigid body by combining a net force with a free moment (or couple). The net effect on the body can be described by both its translation (due to the force) and rotation (due to the moment), and these two effects can be analyzed together.

Moment (Torque)

The moment, or torque, is a measure of the tendency of a force to rotate an object about a pivot or an axis. It is calculated as the product of the force magnitude and the perpendicular distance from the force’s line of action to the pivot point, and it is pivotal in understanding rotational equilibrium in systems.

Equivalence of Force-Couple Systems

This concept involves the principle that a force-couple system can be shifted to a different reference point while maintaining the same external effect on a body. When relocating the system, the force remains unchanged but the moment is adjusted by taking into account the moment generated by the force’s new position relative to the original point.

Transcript

0:00 Problem 3 .96.

00:01 In this problem we have to find equivalent force at point c and equivalent moment at the same point while the wooden stick is applying 175 newton force on the door.

00:17 Now this door is hinged about point c so this door can rotate about point c so this wooden stick is preventing its rotational motion.

00:30 So this will apply a force, let's say in the direction a to b.

00:36 So first of all we will write the p force in its vector form and for that we need the coordinate of point a and point b.

00:46 So if you write the coordinate of point a, so it will be since in x direction it is at minus 100 mm so it will be if we write it in meter so minus 0 .1 and in it is in xxx plane so its y coordinate will be zero and its z coordinate is 594 m m so 0 .594 meter now talking about the coordinates of point b since point b is in again in the negative x direction but at a distance of 67 mm so it will be minus 0 .067 its y coordinate will be 990 mm so 0 .99 and its z coordinate since it is in xy plane so its z coordinate is going to be 0 now talking about the coordinate of point c so it will be in x direction it is at minus 750 m m so minus 0 .75 its y coordinate is 1850 so it will be 1 .85 and its z coordinate will be zero since it is in x y plane now talking about the force that is acting along the wooden stick ab so reaction or the equivalent moment equivalent force at point c will be seen the force that is acting along the stick ab since only one force is present in the system and we can write this force should be since its moment is its magnitude is given as 175 newton so it will be 175 lambda ab where lambda ab is the unit vector along the odinistic ab and it can be written as 175 ab vector divided by mod of ab vector now coordinates of point a and point b is already known so we can solve this so if we write the ab vector it comes out to be 175 033 i cap and point plus 099 j cap and minus 0 .59 kx and minus 0 .59 kx cap divided by its magnitude and magnitude is under root of 033 square plus 0 .99 full square plus minus 0 .59 4 square so if you solve this it comes out to be 5 i cap plus 150 j cap minus 90 k cap all the units are in newton since this is the expression for force now talking about the equivalent moment at point c for that we can write we can see here that a position vector will be this that will be position vector of point b with respect to point c so moment about point c can be calculated by using position vector of point b with respect to c cross multiplied by force that is acting at point along the line ab and we have already calculated the force acting along the line ab and since rc we asked the equals to b so we can write now we need to calculate the position vector of b with respect to c now the coordinates of b and is already known so we can solve this and the position vector comes out to be 0 .683 icap minus 0 .86 jcap and the k -cap both have the same value so there will be no k -cap component this will be cross multiplied by the force that is written here and that will be 5 i -cap and that will be 5 i -cap plus 150 j cap minus 90 k cap we can solve this by using determinant method so it will be i cap j cap now coefficient of i cap is 0 .683 coefficient of j cap is minus 0 .86 k cap coefficient is 0 and in the force vector this is 5 this is plus 150 and this is minus 90 so now now we can expand this expression along icap, jcap and k cap...

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