a uniform beam 6.0m long and weighing 4kg rest on support at P & Q placed left and right 1.0m from each end of the beam. weight of mass 10kg and 8kg are placed near P & Q respectively on each end of the beam. calculate the reaction of P & Q.
pls solve it
To calculate the reactions at supports P and Q, we can consider the equilibrium of moments and forces acting on the beam.
1. Calculate the total weight of the beam:
The weight of the beam is given as 4 kg.
2. Calculate the total weight on each end of the beam:
For the left end near P, there is an additional weight of 10 kg.
For the right end near Q, there is an additional weight of 8 kg.
The weight on each end is 10 kg + 4 kg = 14 kg near P, and 8 kg + 4 kg = 12 kg near Q.
3. Calculate the distance of each weight from the supports:
The weight near P is placed 1.0 m from the left end, so the distance from support P is 1.0 m.
The weight near Q is placed 1.0 m from the right end, so the distance from support Q is 1.0 m.
4. Calculate the reaction at support P:
Taking the moments about support P, we can write:
(14 kg x 9.8 m/s^2 x 1.0 m) - (4 kg x 9.8 m/s^2 x 3.0 m) = 0
(137.2 N*m) - (117.6 N*m) = 0
The total moment about support P is equal to zero.
The reaction at support P is the sum of the weight of the beam and the weight on the left end of the beam:
Reaction at P = 4 kg x 9.8 m/s^2 + 10 kg x 9.8 m/s^2 = 39.2 N + 98 N = 137.2 N
5. Calculate the reaction at support Q:
Taking the moments about support Q, we can write:
(12 kg x 9.8 m/s^2 x 1.0 m) - (4 kg x 9.8 m/s^2 x 3.0 m) = 0
(117.6 N*m) - (117.6 N*m) = 0
The total moment about support Q is equal to zero.
The reaction at support Q is the sum of the weight of the beam and the weight on the right end of the beam:
Reaction at Q = 4 kg x 9.8 m/s^2 + 8 kg x 9.8 m/s^2 = 39.2 N + 78.4 N = 117.6 N
Therefore, the reaction at support P is 137.2 N and the reaction at support Q is 117.6 N.
To calculate the reactions at points P and Q, we need to find the sum of moments and the sum of forces acting on the beam.
Let's start by analyzing the forces acting on the beam.
1. Weight of the beam (4 kg): The weight acts at the center of the beam, so it is evenly distributed between P and Q. Therefore, each support will experience half of the weight, which is (4 kg * 9.8 m/s^2) / 2 = 19.6 N.
2. Weight of the mass near P (10 kg): The weight acts downward at point P.
3. Weight of the mass near Q (8 kg): The weight acts downward at point Q.
Now, let's consider the moments about point P to solve for the reaction at Q.
Summing the moments about P:
Clockwise moments:
Moment due to the beam = 19.6 N * 3 m = 58.8 Nm (since P is 1m away from the center of the beam on the left side)
Counter-clockwise moments:
Moment due to the 10 kg mass = 10 kg * 9.8 m/s^2 * 5 m = 490 Nm (since P is 1m away from the 10 kg mass on the left side)
To be in equilibrium, the total sum of moments must be zero:
Total sum of moments = 58.8 Nm - 490 Nm = -431.2 Nm
Next, let's solve for the reaction at Q by summing the forces in the vertical direction:
Summing the forces at Q:
Upward forces:
Reaction at Q = 19.6 N (due to the weight of the beam)
Downward forces:
Weight of the 8 kg mass = 8 kg * 9.8 m/s^2 = 78.4 N
To be in equilibrium, the sum of forces must be zero:
Sum of forces = Reaction at Q - Weight of the 8 kg mass = 0
Therefore, Reaction at Q = Weight of the 8 kg mass = 78.4 N
To summarize:
Reaction at P = 19.6 N
Reaction at Q = 78.4 N