A Uniform beam 6.0 metre long and weighing 4kg rest 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
The cg of the beam itself is at the 3.0m mark
Summing moments about P (the 10kg end). P,Q are 4 m apart.
4*8 -4Q+2*4=0
Q=8+2=10 *9.8 Newtons
Summing moments about Q:
4*10-4*P+2*4=0
P= 2+10=12*9.8 Newtons
check P+Q=4+10+8=22*9.8N
Well, I must say, it sounds like quite the balancing act! Let's see if we can calculate those reactions for you.
To find the reactions at supports P and Q, we need to consider the equilibrium of moments. Since the beam is at rest, the sum of the moments about any point must be zero.
Let's start by calculating the clockwise moments acting about point P:
Clockwise moment = weight * distance
Moment at P = 10kg * 1.0m
Now, let's calculate the anticlockwise moments acting about point P:
Anticlockwise moment = reaction at Q * distance
Moment at P = reaction at Q * (4.0m - 1.0m)
Since the beam is in equilibrium, the sum of these moments must be zero:
0 = (10kg * 1.0m) - (reaction at Q * (4.0m - 1.0m))
Simplifying this equation, we find:
reaction at Q = (10kg * 1.0m) / (4.0m - 1.0m)
Once we calculate the reaction at Q, we can find the reaction at P by subtracting it from the total weight of the beam. Given that the beam weighs 4kg and there is an additional 8kg weight at the end:
reaction at P = (4kg + 8kg) - reaction at Q
So, let's crunch the numbers and calculate those reactions! Just remember to perform the calculations carefully, unlike a clumsy clown like me.
To calculate the reactions at supports P and Q, we need to consider the equilibrium of the beam.
Step 1: Calculate the total weight of the beam, which is given as 4 kg. Convert it to Newtons by multiplying it with the acceleration due to gravity (9.8 m/s^2):
Total weight of the beam = 4 kg * 9.8 m/s^2 = 39.2 N
Step 2: Calculate the total weight due to the 10 kg mass near support P. Since it is 1 meter away from the support, it will have a moment (torque) acting on support P. The total weight is given by:
Weight due to the 10 kg mass = 10 kg * 9.8 m/s^2 = 98 N
Step 3: Calculate the total weight due to the 8 kg mass near support Q. Again, since it is 1 meter away from the support, it will have a moment (torque) acting on support Q. The total weight is given by:
Weight due to the 8 kg mass = 8 kg * 9.8 m/s^2 = 78.4 N
Step 4: Now, let's assume the reaction force at support P is RP and the reaction force at support Q is RQ. Considering the equilibrium of the beam, we sum up all the forces and moments acting on the beam to be zero.
The sum of vertical forces should be zero:
RP + RQ - Total weight of the beam - Weight due to the 10 kg mass - Weight due to the 8 kg mass = 0
The sum of moments about support P should also be zero:
- RP * 1 m + Weight due to the 10 kg mass * 1 m - Weight due to the 8 kg mass * 5 m = 0
Step 5: Let's solve the equations to find the reactions at supports P and Q:
RP + RQ - 39.2 N - 98 N - 78.4 N = 0 (equation 1)
- RP * 1 m + 98 N * 1 m - 78.4 N * 5 m = 0 (equation 2)
Simplifying equation 1:
RP + RQ = 215.6 N
Simplifying equation 2:
- RP + 98 N - 392 N = 0
- RP = 294 N
RP = -294 N (Note: The negative sign indicates the direction of the force)
Substituting the value of RP in equation 1:
-294 N + RQ = 215.6 N
RQ = 509.6 N
Therefore, the reactions at supports P and Q are:
RP = -294 N
RQ = 509.6 N