a uniform beam 6.0m long and weighing 4kg on support of p and q placed let and right 1.0m from each end of the beam

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To solve this problem, we need to consider the forces acting on the beam and find the reactions at the supports. Let's break down the problem step by step:

Step 1: Draw a diagram
Start by drawing a diagram of the beam with the distances given. Label the beam's length as 6.0m and the distances from the supports p and q as 1.0m each.

Step 2: Identify the forces
Next, identify the forces acting on the beam. Since the beam is at rest and in equilibrium, the sum of the forces and moments acting on it must be zero.

- The weight of the beam (4kg) can be represented by a downward force acting at the center of mass (3.0m). Let's label this force as W.
- The reaction forces at supports p and q can be denoted as Rp and Rq, respectively.

Step 3: Apply the equations of equilibrium
To find the reactions at the supports, we will apply the equations of equilibrium. In statics, equilibrium refers to the condition where all forces and moments acting on an object balance each other out.

In the vertical direction (y-axis), the sum of the forces must be zero:
ΣFy = Rq + Rp - W = 0

In the moment equation, we need to choose a point to calculate the torques. We'll choose the left support (p) as the pivot point, and the clockwise moments as positive:
ΣMp = (W × 3.0) - (Rq × 5.0) = 0

Step 4: Solve the equations
Now we can solve the equations simultaneously to find the reactions Rq and Rp.

From ΣFy:
Rq + Rp - W = 0
Rq + Rp = W

From ΣMp:
(W × 3.0) - (Rq × 5.0) = 0

Substitute the value of W (weight of the beam = 4kg × 9.8m/s^2 = 39.2N):
(39.2N) × 3.0m - (Rq × 5.0m) = 0

From Rq + Rp = W, substitute the value of Rq in terms of Rp:
Rp + (39.2N - Rp) = 39.2N
2Rp = 39.2N
Rp = 39.2N / 2
Rp = 19.6N

Finally, substitute Rp back into the equation Rq + Rp = W:
Rq + 19.6N = 39.2N
Rq = 39.2N - 19.6N
Rq = 19.6N

Therefore, the reaction force at support p is 19.6N, and the reaction force at support q is also 19.6N.

To summarize, the reaction forces at both supports p and q are both 19.6N.