An object is subjected to the forces shown in What single...

Key Concepts

Translational Equilibrium

This concept means that the total sum of all forces acting on the object is zero, so there is no net acceleration. In practice, individual force components along each coordinate axis (such as x and y) are determined and balanced so that the overall force vector is nullified, ensuring the object remains at rest.

Force Decomposition

Force decomposition involves breaking down a force into its component parts along chosen axes (typically x and y axes). This is crucial for analyzing and solving problems in statics, as it allows one to separately balance forces in each direction and simplify calculations of the overall system equilibrium.

Rotational Equilibrium

Rotational equilibrium refers to the condition where the sum of all torques (moments) about any pivot point is zero, meaning the object will not experience angular acceleration. This is important when applying forces at different points since the line of action and the point of application determine the torque produced.

Moment of a Force

The moment (or torque) of a force is the tendency of the force to rotate an object about a pivot point, and is calculated as the product of the force and its perpendicular distance (lever arm) from the pivot. Ensuring that the net moment is zero is essential for achieving rotational equilibrium.

Line of Action

The line of action of a force is an imaginary line along which the force is applied. Choosing the correct point of application, especially on a specific axis like the x-axis, ensures that while the force contributes to balancing the net force, it also counteracts any undesired rotation by aligning appropriately with the moments produced by other forces.

Transcript

00:01 Hello to this question we have to find out a force acting on a data point on xx such that the net system is in equilibrium.

00:08 Let's say this is the force f which is making an angle theta with x axis.

00:14 So if the system is in equilibrium, the net force on the system in x direction and y direction should be equal to 0.

00:21 Now the system of forces given here, this is about 300 newton in x axis, 150 newton in y axis, and 200 newton at an angle of the system.

00:30 Of 30 degrees so component of this force can be taken in x direction and y direction so in case of x direction and here angle is given 40 degree 1 so this angle will be also 40 degree so the net angle with xx is equal to 70 degrees so force component along this axis will be 200 course 70 degree okay and along xxas it will be 200 570 degree so a component of the unknown force f along x -axis and y -axis so along x -axis it will be f -cost theta and along v axis it is f sine theta f -sin -theta so the whole system is in equilibrium then the net force in x direction and y -direction should be equal to zero so summation of all forces in x -direction is equal to zero using this condition we will get in x -direction we have force 300 newton f cost theta and 200 cost 70 degree so minus 300 because 300 acting in negative x -axis plus fcotech plus 200 plus 70 should be equal to 0 so on solving this we get the value of f 31 .6.

02:16 Let's see equation number 1.

02:18 Similarly in y direction, in y direction, net force along y direction must be equal to 0 for the system of 2b in equilibrium.

02:32 Now net force acting in y direction is 150 newton, f sine theta, okay, 200 side 70.

02:39 So 200 side 70 and 150 newton are acting in the same direction.

02:46 So 200 side 70 plus 150 should be equal to f.

02:51 Side theta okay so from here f side theta equal to is approximately equal to 3 6 newton okay let's say equation number two so we have to find out the value of f so squaring first equation and adding with the square of second equation okay so f square sine theta f square v .2x2teta is equal to f squared and f is equal to under root of 3366 plus 231 whole square 336 square plus 231 .6 2 .31 .6 whole square 2 whole power 1 by 2.

03:45 So on solving our f value comes out 410 newton approximately equal to 410 newton okay and we have to also find out the value or direction of this force here we have f cost it equal to 216 and f sain tete tuit equal to 336.

04:07 So according to question this is our f cost theta this is our f sine theta this is our fcata and this is our f sine theta so the result of these two forces equal to f will act along this direction let's say this angle is equal to theta dash so from this diagram we can find out time theta which is equal to perpendicular upon base so perpendicular value is equal to f sine theta divided by fc theta so we have f sine tapp also 336 divided by f post theta that is 132 .6 okay 132 point sorry 231 .6 231 .6.

05:05 So theta is equal to 10 inverse 1 .44 approximate okay so from here theta is equal to sorry theta dash not theta theta dash is equal to 55 .2 degree and this is in negative direction so we can write it equal to minus 55 .2 degree or we can say it is to 360 minus 55, 360 minus 55 .2 degree.

05:44 Okay.

05:46 So this will be equal to 305 approximately value 305 degree.

05:53 Okay? that is this angle is from, let's say, theta dash, from positive x -axis.

06:02 From positive x -axis.

06:06 Okay? so angle may be written as minus 55 .20 degree or 305 from positive x -axis...

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