3 FRICTION Equilibrium of The Body on Ladder

FRICTION ((LADDER PROBLEMS)

MAIT

Q. 1: What is ladder friction? How many forces are acting on a ladder? Solution: A ladder is an arrangement used for climbing on the walls It essentially consists of two long uprights of wood or iron and connected by a number of cross bars. These cross bars are called rungs and provide steps for climbing. Fig shows a ladder AB with its end A resting on the ground and end B leaning against a wall. ll. The ladder is acted upon by the following set of forces: (1) Weight W acting downwards at its mid point. (2) Normal reaction Rb and friction force Fb = µRb at the end B leaning against the wall. Since the ladder has a tendency to slip downwards, the friction force will be acting upwards. If the wall is smooth (µ = 0), the friction force will be zero. (3) Normal reaction Ra and friction force Fa =µRa at the end A resting on the floor. Since the ladder, upon slipping, tends to move away from the wall, the direction of friction force will be towards the wall. Applying equilibrium conditions, the algebraic sum of the horizontal and vertical component of forces would be zero.

Q. 2: A ladder 5m long rests on a horizontal ground and leans against a smooth vertical wall at an angle 70° with the horizontal. The weight of the ladder is 900N and acts at its middle. The ladder is at the point of sliding, when a man weighing 750N stands 1.5m from the bottom of the ladder. Calculate coefficient of friction between the ladder and the floor. Solution: Forces acting on the ladder is shown in fig 9.62 Resolving all the forces vertically, RV = R – 900 – 750 = 0 R = 1650N ...(ii) Now taking moment about point B, R × 5sin20° – Fr × 5cos20° – 900 × 2.5sin20° – 750 × 3.5sin20° = 0 Since Fr = µR, and R = 1650N; Fr = 1650µ Putting, the value of R and Fr µ = 0.127 .......ANS (see Fig.)

Q. 3: A uniform ladder of length 13m and weighing 250N is placed against a smooth vertical wall with its lower end 5m from the wall. The coefficient of friction between the ladder and floor is 0.3. Show that the ladder will remain in equilibrium in this position. What is the frictional force acting on the ladder at the point of contact between the ladder and the floor? Solution: Since the ladder is placed against a smooth vertical wall, therefore there will be no friction at the point of contact between the ladder and wall.

Compiled by: RAMAKANT RANA “[email protected]”

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FRICTION ((LADDER PROBLEMS)

MAIT

Resolve all the forces horizontally and vertically. ∑H = 0, ...(i) Fr – R2 = 0 ∑V = 0, ...(ii) R1 – 250 From the geometry of the figure, BC = 12m Taking moment about point B, R1 × 5 – Fr × 12 – 250 × 2.5 = 0 Fr = 52N .......ANS For equilibrium of the ladder, Maximum force of friction available at the point of contact between the ladder and the floor = µR = 0.3 × 250 = 75N .......ANS Thus we see that the amount of the force of friction available at the point of contact (75N) is more than force of friction required for equilibrium (52N). Therefore, the ladder will remain in equilibrium in this position.

Q. 4: A uniform ladder of 7m rests against a vertical wall with which it makes an angle of 45º, the coefficient of friction between the ladder and the wall is 0.4 and that between ladder and the floor is 0.5. If a man, whose weight is one half of that of the ladder, ascends it, how high will it be when the ladder slips? Solution: Let, X = Distance between A and the man, when the ladder is at the point of slipping. W = Weight of the ladder Weight of man = W/2 = 0.5W Fr1 = 0.5R1 ...(i) Fr2 = 0.4R2 ...(ii) Resolving the forces vertically R1 + Fr2 – W – 0.5W = 0 R1 + 0.4R2 = 1.5W ...(iii) Resolving the forces Horizontally R1 ...(iv) R2 – Fr1 = 0; R2 = 0.5R Solving equation (iii) and (iv), we get 0.25W R2 = 0.625W, Fr2 = 0.25 Now taking moment about point A, W × xcos45° – R2 × W × 3.5cos45° + 0.5W 7sin45° – Fr2 × 7cos45° Putting the value of R2 and Fr2, we get .......ANS X = 5.25m

Compiled by: RAMAKANT RANA “[email protected]”

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