A uniform beam 6.0 metres long and weighing 4 kg rests on supports at P and Q placed left and right 1.0 metre from each end of the beam. weights of mass 10kg and 8kg are placed near P and Q respectively one each end of the beam. calculate the reactions at P and Q?
To calculate the reactions at points P and Q, we need to consider the equilibrium of the uniform beam. The beam is in equilibrium when the sum of the forces acting on it is zero, and the sum of the moments (torques) acting on it is also zero.
Let's start by calculating the reactions at points P and Q separately.
Step 1: Calculate the total weight of the beam.
The beam itself weighs 4 kg. We can convert this weight into Newtons by multiplying it by the acceleration due to gravity, which is approximately 9.8 m/s^2.
Weight of the beam = mass x acceleration due to gravity
Weight of the beam = 4 kg x 9.8 m/s^2 = 39.2 N
Step 2: Calculate the total weight acting on point P.
There is a weight of 10 kg placed near point P. The distance between point P and the 10 kg weight is 1 meter.
Weight on P = weight of the beam + weight of the 10 kg weight
Weight on P = 39.2 N + (10 kg x 9.8 m/s^2) = 39.2 N + 98 N = 137.2 N
Step 3: Calculate the total weight acting on point Q.
There is a weight of 8 kg placed near point Q. The distance between point Q and the 8 kg weight is also 1 meter.
Weight on Q = weight of the beam + weight of the 8 kg weight
Weight on Q = 39.2 N + (8 kg x 9.8 m/s^2) = 39.2 N + 78.4 N = 117.6 N
Step 4: Calculate the reactions at points P and Q.
Since the beam is in equilibrium, the sum of the forces and the sum of the moments acting on it are both zero.
Sum of the forces in the vertical direction = R_P + R_Q - Weight of the beam - Weight of the 10 kg weight - Weight of the 8 kg weight
0 = R_P + R_Q - 39.2 N - 98 N - 78.4 N
R_P + R_Q = 215.6 N
Sum of the moments clockwise (taking point P as the pivot) = (1 m) * (Weight on Q) - (5 m) * (Weight on P) - (6 m) * (Weight of the beam)
0 = (1 m) * (117.6 N) - (5 m) * (137.2 N) - (6 m) * (39.2 N)
0 = 117.6 N - 686 N - 235.2 N
0 = -803.6 N + 117.6 N
Rearranging the equation R_P + R_Q = 215.6 N gives us R_P = 215.6 N - R_Q.
Substituting this value into the equation above, we have 0 = -803.6 N + 117.6 N + R_Q.
Simplifying, we get R_Q = 686 N - R_P.
Now we have two equations:
R_P + R_Q = 215.6 N
R_P = 215.6 N - R_Q
R_Q = 686 N - R_P
We can solve these equations simultaneously to find the values of R_P and R_Q.