A uniform ladder 3 m long weighing 20 kg leans against a frictionless wall. Its foot rest on a

A uniform ladder 3 m long weighing 20 kg leans against a...

Key Concepts

Static Equilibrium

Static equilibrium refers to the condition where an object remains at rest, meaning that the sum of all forces and the sum of all moments (torques) acting on the object are zero. This concept is fundamental in analyzing problems in mechanics where no net motion occurs.

Torque (Moment) Equilibrium

Torque equilibrium involves analyzing the rotational forces (moments) acting about a pivot point to ensure that the net torque is zero. This is essential in determining how forces such as weight and reaction forces balance in a system, especially when the forces act at different distances from the pivot.

Force Decomposition

Force decomposition is the process of breaking down forces into their horizontal and vertical components. This technique simplifies the analysis of forces acting on a body, making it easier to apply the conditions of static equilibrium in each direction.

Reaction Forces

Reaction forces are the forces exerted by surfaces to support other bodies in contact with them. In problems involving contact with rigid surfaces, it is important to account for the normal and frictional components of these forces, which are critical in maintaining the equilibrium of the system.

Center of Mass

The center of mass is the point at which the mass of an object is considered to be concentrated. In uniformly distributed bodies, such as a uniform ladder, the center of mass is located at the midpoint. This concept is crucial for analyzing how gravitational force acts on extended bodies in equilibrium problems.

Transcript

00:01 In this question we have given a uniform ladder of length 3 meter and weight 20 kilograms which rest against a frictionless wall we have to calculate the reaction force by the wall and the floor so first of all i am drawing the diagram for this given cushion so this is the frictionless wall and this is a floor and there is a ladder so this is our ladder which is resting i'm going to show the forces acting on this ladder so one force is the reaction force by this wall that is acting in this direction let this is r1 and one is the normal force by this floor that is in vertical direction this is equals to n and one is the frictional force between the ladder and this floor that is f so we can see that the resultant of this both the force is this is given by that is r2 so this is our reaction force r2 weight of this ladder is acting vertically downward weight is 20 g and this distance is given to us that is 1 meter and this distance is given to us that is length of the ladder that is 3 meter so we can say this is equal to 2 root 2 meter from pythagoras now let this point is a this is the gravity center of gravity and this is point b so we can say that for translatory equilibrium net horizontal force should be 0 so we can say that for translational equilibrium net horizontal force should be 0 so we can say that this r1 force minus the frictional force should be 0 so from here we can say that f is equals to this is r1 let this is our equation first now for we can say that vertical in vertical force should be 0 so net vertical force should also be 0 for translatory equilibrium so we can say that n should be equals to weight of the ladder that is 20 into 10 that is 200 newton this is our equation second this is the value of normal force now we can say for rotational equilibrium i'm going to take moment about center of gravity or we say take moment about point a that should be 0 so we can say that r1 into this is 2 root 2 minus moment due to the weight that is 20g into this is 1 by 2 is equal to 0 so we can say this will come out to be from here r1 and that is equals to 200 upon this is 4 root 2 so we can say this is equal to 25 root 2 newton so this is the r1 force and this is the reaction force by the wall and we can say that from equation to f should be equals to r1 that is 25 root 2 newton.

03:24 We have to calculate the value of r2 and this is equal to under root of n square plus f square.

03:31 Now put all the data in this equation.

03:33 So this is under root of 200 square plus 25 root 2 square...

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