A rod 2.5m long is supported at A and B. The rod is carrying a load of 5kN at a distance of 1m from A. What are the reactions at A and B?
To find the reactions at supports A and B, we will use the principles of equilibrium. The sum of the vertical forces and the sum of the moments about any point must be equal to zero.
Let's start by calculating the reactions at support A and support B.
1. Sum of vertical forces:
The vertical forces acting on the rod are the load at a distance of 1m from support A and the reactions at A and B. Since the rod is in equilibrium, the sum of the vertical forces must be zero.
ΣFy = 0
2. Sum of moments about point A:
The moment of a force is the product of the force and the perpendicular distance from the point of rotation. Taking moments about point A, the distance of the 5kN load from A is 1m.
Applying the principle of moments, the sum of the moments about point A must be zero.
ΣM A = 0
From these two equations, we can determine the reactions at supports A and B.
Let's solve the equations step-by-step:
1. Sum of vertical forces:
Taking upward forces as positive,
+R A - 5kN - R B = 0
R A - R B = 5kN
2. Sum of moments about point A:
Taking clockwise moments as positive,
-5kN * 1m - R B * 2.5m = 0
-5kN - 2.5m * R B = 0
-2.5m * R B = 5kN
R B = -5kN / -2.5m
R B = 2kN
Now, substituting the value of R B into the first equation:
R A - 2kN = 5kN
R A = 5kN + 2kN
R A = 7kN
Therefore, the reactions at support A and support B are 7kN and 2kN, respectively.