A uniform 5.0-kg ladder is leaning against a frictionless vertical wall, with which it makes a 1

A uniform 5.0-kg ladder is leaning against a frictionless...

Key Concepts

Static Equilibrium

Static equilibrium is the state in which a system experiences no net force or net torque, meaning that every force or moment acting on it is balanced by an equal and opposite force or moment. This concept is fundamental when analyzing objects like ladders, beams, or structures that must remain stationary despite applied forces.

Torque and Rotational Equilibrium

Torque is the measure of the force that causes an object to rotate about an axis or pivot. Rotational equilibrium is achieved when the sum of all torques acting on a system is zero, ensuring that the system does not rotate. In problems involving ladders or beams, torques created by the weight of the ladder, a person, and other forces must be balanced to prevent rotation.

Friction and Coefficient of Friction

Friction is the resistive force that occurs when two surfaces interact, and the coefficient of friction quantifies the amount of frictional resistance. It is crucial in preventing motion, particularly in systems where one surface is frictionless and another provides resistance. Understanding friction is essential for determining the conditions under which an object like a ladder will remain stationary or begin to slip.

Center of Mass

The center of mass is the point in a body or system where its mass is considered to be concentrated. For uniform objects, such as a ladder, the center of mass is located at the geometric center. This concept is essential in analyzing the forces and torques acting on the ladder, as the weight of the ladder exerts its force through its center of mass.

Transcript

00:01 In the given problem, there is a ladder which is put against a vertical wall making an angle of just 15 degree with this wall.

00:18 If this angle is 15 degrees, then obviously this angle will be 75 degree as the angle between ground and the wall is 90 degree.

00:30 As this letter is uniform so its weight will be acting at its center of mass as its mass is 5 kilogram so its weight will be 5g let the letter be named as a b its length can be taken as l so it's central point c if we mark it here then this ac will be l by suppose in first case the person manages to reach up to this uppermost point of the letter.

01:17 The weight of the person 65g will be acting here at the point b.

01:23 Now the normal reaction of this end a of the ladder over the ground is normal reaction.

01:34 By the ground on this end of the ladder, that is n -a.

01:39 And normal reaction exerted by the vertical wall at the other end of the ladder is n as the ladder is in equilibrium.

01:50 So summation of torque should be zero acting on it, and summation of all the forces acting on it should also be zero.

02:00 Hence we can see all the vertical forces should be zero, the algebraic sum of all the vertical forces and all the horizontal forces should be zero.

02:12 This is so as all the vertical forces are coming to be zero hence so as vertical forces are zero hence vertically here this is an a vertically up and 5g and 65g are vertically down so na will be equal to 5g plus 65g means it will come out to be 70 g.

02:40 And as far as horizontal forces are concerned, as the tendency of this end, a, of the letter, will be to slide leftward over the ground.

02:50 So there will be a force of friction acting on it, which will keep it in its position.

02:59 So that is acting leftward.

03:02 So using that, nb here is equal to f and we know force of friction is given as mu into normal reaction and here normal reaction is this n a so finally this reaction of the vertical wall on the ladder comes out to be mu times of n a or we can say mu times of 70 g taking the equilibrium of this ladder against the rotation and taking the rotation about point a.

03:39 So we will find out the torques about point a, counterclockwise torque and clockwise torque.

03:45 And we will look for their values.

03:49 So first of all, we find clockwise torque about point a of the latter.

04:06 And it comes out to be 65g into its perpendicular distance from the.

04:17 Point of rotation means this is 65g so its vertical distance will be this let it be d so this is a d this is so that is ab cost 75 degree hence it becomes 65g into ab and for ab we can use l length of the letter l cause 75 degree then the torque produced by the mass of the letter itself bit of the letter itself, that is 5g, and this will be l by 2 into cost 75 degree.

04:54 As we look for the perpendicular distance of this line of action of 5g, this is the perpendicular distance here this.

05:02 So for this, we will use ac, which is l by 2.

05:06 That's why.

05:09 So now calculating it, here it comes out to be 164 .87l.

05:17 And here it comes out to be 6 .34 into l.

05:23 So adding them, it becomes 171 .21 l newton into meter...

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