A plank AB 3m long weighing 20kg with its centre of gravity 2m from the end A carries a load of mass 10kg at the end A. it rests on two supports at C and D
1. compute the values of the reaction forces R1 and R2 at C and D.
2. How far from D, and on what side of it, must a mass 24kg be placed on the plank so as to make the reactions equal ? What are their values ?
3. without this 24kg, what vertical force applied at B will just lift the plank clear of D? What is then the reaction at C?
To solve the given problem, we will use the principles of static equilibrium.
1. Computing the values of the reaction forces R1 and R2 at C and D:
Let's start by drawing a free body diagram of the plank AB:
A --------- B
/ /
C D
At the point C, there will be a vertical reaction force R1 acting upwards. At the point D, there will be a vertical reaction force R2 acting upwards as well.
To find the values of R1 and R2, we need to analyze the vertical and horizontal forces acting on the plank. The sum of vertical forces must be zero for the plank to be in equilibrium.
Vertical forces:
- Weight of the plank (20 kg) acting downwards at the center of gravity (2 m from A). This force is given by F1 = m1 * g, where m1 = 20 kg and g = 9.8 m/s^2.
- Weight of the load (10 kg) at end A acting downwards. This force is given by F2 = m2 * g, where m2 = 10 kg.
Considering the vertical equilibrium, we have:
R1 + R2 = F1 + F2
Substituting the values:
R1 + R2 = (20 kg * 9.8 m/s^2) + (10 kg * 9.8 m/s^2)
R1 + R2 = (196 N) + (98 N)
R1 + R2 = 294 N
Since the plank is symmetrical, we can conclude that R1 = R2 = 294 N.
Therefore, the values of the reaction forces R1 and R2 at C and D are both 294 N.
2. Placing a mass of 24 kg on the plank to make the reactions equal:
To make the reactions at C and D equal, we need to find the location on the plank where the 24 kg mass should be placed. Let's denote this distance as 'x' from point D.
Considering the moments (torques) about D, the sum of moments must be zero for the plank to be in equilibrium.
Moment about D:
Clockwise moment = Distance from D to center of gravity (2 m) * Weight of the plank (20 kg) = 2 m * 20 kg * 9.8 m/s^2
Counterclockwise moment = Distance from D to the new mass (x) * Weight of the added mass (24 kg) = (3 m - x) * 24 kg * 9.8 m/s^2
Equating the moments:
2 m * 20 kg * 9.8 m/s^2 = (3 m - x) * 24 kg * 9.8 m/s^2
Simplifying:
40 m * 9.8 m/s^2 = (3 m - x) * 24 kg * 9.8 m/s^2
x = (3 m * 24 kg) / 40 kg
x = 1.8 m
The 24 kg mass should be placed at a distance of 1.8 m from point D, on the opposite side of D compared to A.
3. Finding the vertical force required to lift the plank clear of D:
If we remove the 24 kg mass from the previous case, we need to find the vertical force applied at B that will just lift the plank clear of D. Let's denote this force as 'F'.
Considering the vertical equilibrium, we have:
F = Weight of the plank (20 kg) + Weight of the load at A (10 kg) + Reaction force at D (R2)
Substituting the values:
F = (20 kg * 9.8 m/s^2) + (10 kg * 9.8 m/s^2) + 294 N
F = (196 N) + (98 N) + 294 N
F = 588 N
Therefore, the vertical force applied at B to lift the plank clear of D is 588 N. The reaction at C remains the same as in the first case: R1 = 294 N.