A ladder weighing 400N is 10m long. Its end ‘A’ is resting on smooth vertical wall and lower end ‘B’ is resting on rough horizontal floor having coefficient of friction is 0.4. The ladder makes an angle 45o with horizontal. The ladder is about to slip when a man weighing 600N standing at mid length of the ladder. Find reactions at supports A and B and limiting friction at the floor.
Good
To find the reactions at supports A and B and the limiting friction at the floor, we can use the principles of equilibrium.
Step 1: Resolve the forces acting on the ladder.
- The weight of the ladder, Wladder = 400N, can be divided into horizontal and vertical components.
- The horizontal component is Wladder * sin(45) = 400 * sin(45) = 282.84N.
- The vertical component is Wladder * cos(45) = 400 * cos(45) = 282.84N.
Step 2: Resolve the forces acting on the man.
- The weight of the man, Wman = 600N, can also be divided into horizontal and vertical components.
- The horizontal component is Wman * sin(45) = 600 * sin(45) = 424.26N.
- The vertical component is Wman * cos(45) = 600 * cos(45) = 424.26N.
Step 3: Calculate the vertical forces.
- The vertical forces acting on the ladder are the vertical component of the ladder's weight (282.84N) and the vertical component of the man's weight (424.26N).
- Therefore, the total vertical force is 282.84 + 424.26 = 707.10N.
Step 4: Calculate the horizontal forces.
- The horizontal forces acting on the ladder are the horizontal component of the ladder's weight (282.84N) and the horizontal component of the man's weight (424.26N).
- Therefore, the total horizontal force is 282.84 + 424.26 = 707.10N.
Step 5: Calculate the reaction at support A.
- The reaction at support A is equal to the total horizontal force acting on the ladder, which is 707.10N.
Step 6: Calculate the reaction at support B.
- The reaction at support B is equal to the total vertical force acting on the ladder, which is 707.10N.
Step 7: Calculate the limiting friction at the floor.
- The limiting friction is given by the equation, limiting friction = coefficient of friction * normal force.
- The normal force is equal to the vertical component of the ladder's weight plus the vertical component of the man's weight, which is 707.10N.
- Therefore, the limiting friction = 0.4 * 707.10 = 282.84N.
Summary:
- The reaction at support A is 707.10N.
- The reaction at support B is 707.10N.
- The limiting friction at the floor is 282.84N.
To find the reactions at supports A and B, as well as the limiting friction at the floor, we can start by analyzing the forces acting on the ladder.
1. Weight of the ladder:
The weight of the ladder is given as 400N and acts vertically downward from its center of mass. This force can be represented as F_ladder = 400N.
2. Weight of the man:
The weight of the man is given as 600N and acts vertically downward from the mid-point of the ladder. Since he is standing at the mid-length of the ladder, the distance of the weight from support A is 5m and from support B is also 5m. This force can be represented as F_man = 600N.
3. Normal reactions at supports A and B:
Since the ladder is in equilibrium, the sum of vertical forces must be zero. The normal reaction at support A, R_A, cancels out the vertical component of F_ladder and F_man. The normal reaction at support B, R_B, cancels out the remaining vertical component of F_ladder. Therefore, we have:
R_A + R_B - F_ladder - F_man = 0.
4. Frictional force at support B:
The ladder is about to slip, which means the frictional force at support B is at its maximum value. The frictional force can be calculated using the equation F_friction = μ * R_B, where μ is the coefficient of friction (given as 0.4).
Now, using the given information and the equations mentioned above, we can calculate the values of R_A, R_B, and F_friction.
1. Calculate R_A:
R_A = F_ladder + F_man - R_B
= 400N + 600N - R_B.
2. Calculate R_B:
R_B = (F_ladder + F_man) / 2
= (400N + 600N) / 2
= 1000N / 2
= 500N.
3. Calculate F_friction:
F_friction = μ * R_B
= 0.4 * 500N
= 200N.
To summarize:
- The reaction at support A (R_A) is equal to 400N + 600N - R_B.
- The reaction at support B (R_B) is equal to 500N.
- The limiting friction at the floor (F_friction) is equal to 200N.