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?
1. The reaction forces R1 and R2 at C and D can be computed using the equation of equilibrium:
R1 + R2 = 20kg + 10kg = 30kg
R1 = 15kg
R2 = 15kg
2. The 24kg mass must be placed 1m from D on the side of C. The reaction forces will then be equal and their values will be 18kg.
3. The vertical force applied at B to lift the plank clear of D is 30kg. The reaction at C will then be 0kg.
1. While considering the static equilibrium of the plank and the load:
Let's assume the reaction force at support C is R1, and the reaction force at support D is R2.
Taking moments about support D:
(3m)*(20kg) + (2m)*(10kg) - R1*(3m) = 0
Simplifying the equation:
60kg + 20kg - 3mR1 = 0
80kg = 3mR1
R1 = (80kg) / (3m)
R1 ≈ 26.67kg
Taking moments about support C:
(3m)*(20kg) + (3m - 2m)*(10kg) - R2*(3m) = 0
Simplifying the equation:
60kg + 10kg - 3mR2 = 0
70kg = 3mR2
R2 = (70kg) / (3m)
R2 ≈ 23.33kg
Therefore, the values of the reaction forces R1 and R2 at supports C and D, respectively, are approximately 26.67kg and 23.33kg.
2. To make the reactions at both supports equal, let's assume the distance from D to the new load of 24kg is x meters.
Taking moments about support D:
(3m)*(20kg) + (2m)*(10kg) - R1*(3m) - (x)(24kg) = 0
Simplifying the equation:
60kg + 20kg - 3mR1 - 24kgx = 0
80kg - 24kgx = 3mR1
Since R1 is equal to R2 (from the previous question), we can write:
80kg - 24kgx = 3mR2
80kg - 24kgx = 3mR1
Substituting the values of R1 and R2:
80kg - 24kgx = 3m(26.67kg)
80kg - 24kgx = 80.01kg
-24kgx = 0.01kg
x ≈ -0.00042m
As the distance cannot be negative, we take the magnitude:
x ≈ 0.00042m
Since the distance is very small and positive, we can conclude that the mass of 24kg must be placed on the plank approximately 0.00042m away from the support D.
3. To calculate the vertical force applied at B that will just lift the plank clear of D, we need to find the reaction at C.
Taking moments about support D:
(3m)*(20kg) + (2m)*(10kg) - R1*(3m) - (3m - 2m)(F_B) = 0
Simplifying the equation:
60kg + 20kg - 3mR1 - (m)(F_B) = 0
80kg - 3mR1 = mF_B
Since the plank will just lift clear of D, the reaction force at D will be zero, so R2 = 0.
Taking moments about support C:
(3m)*(20kg) + (3m - 2m)*(10kg) + R2*(3m) - (m)(F_B) = 0
Simplifying the equation:
60kg + 10kg + 3mR2 - mF_B = 0
70kg + 3mR2 = mF_B
Since R2 = 0:
70kg = mF_B
Therefore, the vertical force applied at B that will just lift the plank clear of D is 70kg, and the reaction force at C is zero.
To solve these problems, we'll use the principles of static equilibrium. For an object to be in equilibrium, the sum of the forces acting on it must be zero, and the sum of the torques about any point must also be zero.
1. Let's start by analyzing the forces acting on the plank AB when it carries a load of 10kg at the end A.
Considering the forces in the vertical direction, we have:
R1 + R2 - 10kg - 20kg = 0 (sum of vertical forces equals zero)
Considering the torques about point C, we have:
(R2 * 3m) + (10kg * 1m) - (20kg * 2m) = 0 (sum of torques equals zero)
Solving these equations will give us the values of R1 and R2.
From the first equation:
R1 + R2 = 30kg
From the second equation:
3R2 + 10kg - 40kg = 0
3R2 = 30kg
R2 = 10kg
Substituting the value of R2 in the first equation:
R1 + 10kg = 30kg
R1 = 20kg
So, the values of the reaction forces are R1 = 20kg and R2 = 10kg.
2. Now, let's find the location and side where a 24kg mass should be placed to make the reactions equal.
Considering the forces in the vertical direction, we have:
R1 + R2 - 24kg - 20kg = 0 (sum of vertical forces equals zero)
Solving this equation will give us the value of R1 + R2.
R1 + R2 = 44kg
Since we want to make R1 equal to R2, we can divide the sum by 2:
(R1 + R2) / 2 = 22kg
This means that each reaction force should be 22kg.
Considering the torques about point D, we have:
(R2 * 3m) + (24kg * x) - (20kg * 2m) = 0 (sum of torques equals zero)
Solving this equation will give us the location x.
3R2 + 24kg * x - 40kg = 0
3R2 + 24kg * x = 40kg
10kg + 24kg * x = 40kg
24kg * x = 30kg
x = 1.25m
So, a mass of 24kg should be placed on the plank 1.25m from point D on the side opposite to point C.
The values of the reactions will be R1 = R2 = 22kg.
3. Now, let's find the vertical force applied at point B that will just lift the plank clear of D.
Considering the torques about point D, we have:
(R2 * 3m) - (20kg * 2m) + (FB * 1m) = 0 (sum of torques equals zero)
Solving this equation will give us the force FB.
3R2 - 40kg + FB = 0
3R2 = 40kg - FB
FB = 40kg - 3R2
Since we want the force FB to just lift the plank clear of D, its magnitude should be equal to zero.
40kg - 3R2 = 0
3R2 = 40kg
R2 = 40kg / 3
The reaction at C will be equal to R1, which is 22kg.
So, the vertical force applied at point B that will just lift the plank clear of D is FB = 40kg - 3R2, and the reaction at C is R1 = 22kg.
1. To compute the values of the reaction forces R1 and R2 at supports C and D, we need to analyze the forces acting on the plank.
Since the plank is in equilibrium, the sum of all the vertical forces and the sum of all the moments about any point on the plank should be zero.
First, let's consider the vertical forces:
- The weight of the plank itself (20kg) acts vertically downward at its center of gravity, 2m from end A.
- The load of mass 10kg at end A also acts vertically downward.
The vertical forces acting on the plank can be represented by a force downward (W) at the center of gravity, and a force (10g) downward at point A.
Now let's consider the moments about point C:
- The weight of the plank (20kg) creates a clockwise moment.
- The load at point A (10kg) creates an anticlockwise moment.
For the plank to be in equilibrium, the sum of the moments must be zero:
Clockwise moment = (20kg) * (2m) = 40kgm
Anticlockwise moment = (10kg) * (3m) = 30kgm
Since the moments are in equilibrium, the reaction force at point C (R1) should be equal to the weight of the plank (20kg) plus the load at point A (10kg):
R1 = 20kg + 10kg = 30kg
To find the reaction force at point D (R2), we can subtract R1 from the total weight of the plank:
R2 = Total weight of plank - R1
R2 = 20kg - 30kg
R2 = -10kg (negative sign suggests upward force)
Therefore, the value of the reaction force at point C (R1) is 30kg and the value of the reaction force at point D (R2) is -10kg.
2. To make the reactions equal, we need to place a mass of 24kg on the plank. Let's assume this mass is placed at a distance of x meters from D.
Considering the moments about point C again:
Clockwise moment = (20kg) * (2m) = 40kgm
Anticlockwise moment = (10kg) * (3m) + (24kg) * (x) = 30kgm + 24kgx
For the plank to be in equilibrium, the sum of the moments must be zero:
Clockwise moment = Anticlockwise moment
40kgm = 30kgm + 24kgx
Subtracting 30kgm from both sides:
10kgm = 24kgx
Dividing both sides by 24kg:
x = 10kgm / 24kg
Simplifying:
x ≈ 0.42m
Therefore, a mass of 24kg should be placed approximately 0.42m from D, on the opposite side of D.
To compute the values of the reactions (R1 and R2) in this case, we need to consider the moments about point C with the additional mass:
Clockwise moment = (20kg) * (2m) = 40kgm
Anticlockwise moment = (10kg) * (3m) + (24kg) * (0.42m) = 30kgm + 10.08kgm = 40.08kgm
The sum of the moments is still zero, so the values of the reaction forces at points C and D should be equal. Let's call this value R:
R = Total weight of plank + Mass at A
R = 20kg + 10kg + 24kg
R = 54kg
Therefore, the values of the reaction forces (R1 and R2) are approximately 54kg each.
3. Without the additional 24kg mass, to find the vertical force applied at B that would just lift the plank clear of D, we need to consider the moment about point C.
To lift the plank clear of D, the sum of the moments about point C must be zero:
Clockwise moment = (20kg) * (2m) = 40kgm
Anticlockwise moment = Vertical force at B * Distance from C to B
For the plank to be lifted clear of D, Anticlockwise moment = 0. Therefore:
Vertical force at B * Distance from C to B = 40kgm
Vertical force at B * (3m + x) = 40kgm
Rearranging the equation:
Vertical force at B = 40kgm / (3m + x)
Since there is no additional 24kg mass, the reaction at C would be equal to R1, which is 30kg.
Therefore, to find the vertical force applied at B (without the additional mass) that would just lift the plank clear of D, we need to substitute the value of x we found in question 2 (approximately 0.42m) into the equation:
Vertical force at B = 40kgm / (3m + 0.42m)
Simplifying:
Vertical force at B ≈ 8.89kg
Therefore, a vertical force of approximately 8.89kg applied at B would just lift the plank clear of D, and the reaction at C would be approximately 30kg.