A uniform beam PQ of length 4.0m and weight 10N is supported at P and Q. It carries of load of 4N at a point 1.0m from P. What is the reaction at Q.
R=6N
Well, it seems like this beam is quite the balancing act! Let me calculate the reaction at Q, using the power of laughter...just kidding, I'll use some physics instead.
To find the reaction at Q, we need to consider the equilibrium of forces. Since the beam is balanced, the sum of all the vertical forces must be zero.
Let's break it down. We have the weight of the beam (10N) acting downwards at its center (2m from P and 2m from Q). We also have the load (4N) acting downwards 1m from P.
The reaction at Q can be calculated by summing up the vertical forces:
RQ + 10N - 4N = 0
Simplifying, we find:
RQ = 4N - 10N
RQ = -6N
Uh-oh! Looks like the reaction at Q is negative, which means it's acting upwards. So, in a serious (but slightly comical) sense, the reaction at Q is -6N. Clown physics strikes again!
To find the reaction at Q, we need to consider the equilibrium of moments about point P. Let's go step-by-step:
1. Calculate the total clockwise moment about point P:
Moment due to the weight of the beam = Weight of the beam × distance from P to the center of gravity of the beam
= 10 N × 2 m
= 20 Nm
Moment due to the load = Load × distance from P to the load
= 4 N × 1 m
= 4 Nm
2. Calculate the total anticlockwise moment about point P:
Moment due to the reaction at Q = Reaction at Q × distance from P to Q
= Reaction at Q × 4 m
3. Since the beam is in equilibrium (not rotating), the total clockwise moment should be equal to the total anticlockwise moment. Thus:
Total clockwise moment = Total anticlockwise moment
20 Nm + 4 Nm = Reaction at Q × 4 m
4. Simplify the equation:
24 Nm = Reaction at Q × 4 m
5. Solve for the reaction at Q:
Reaction at Q = 24 Nm / 4 m
= 6 N
Therefore, the reaction at point Q is 6 N.
To find the reaction at point Q, we can use the principle of moments.
The principle of moments states that for an object in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
In this case, we can take the point Q as our pivot. Let's consider the forces acting on the beam:
1. The weight of the beam (10N) acting vertically downwards at the midpoint of the beam (2.0m from either end).
2. The weight of the load (4N) acting vertically downwards at a point 1.0m from P (3.0m from Q in the opposite direction).
Since the beam is uniform, we can assume its weight acts through its midpoint (2.0m from each end). The reaction at P will act vertically upwards, and we need to find the reaction at Q, which will also act vertically upwards.
To balance the moments around Q, we need to equate the clockwise and anticlockwise moments.
Clockwise moments:
Moment due to the beam = weight of the beam x distance from Q
= 10N x 2.0m
= 20Nm
Anticlockwise moments:
Moment due to the load = weight of the load x distance from Q
= 4N x 3.0m
= 12Nm
Since the beam is in equilibrium, the sum of the clockwise (20Nm) and anticlockwise (12Nm) moments must be equal.
Therefore, the reaction at Q can be found as the difference between these two moments:
Reaction at Q = Moment due to beam - Moment due to load
= 20Nm - 12Nm
= 8Nm
So, the reaction at Q is 8 newton meters (Nm) in the upward direction.