Question

A uniform plank AB 30 m long, weighing 100 N is pivoted at point P,Q which are 5 m from the ends A and B respectively.A boy of weight 250 N stand at apoint D on the plank,1 m away from Q and the arrangement is in equilibrium.Determine the reaction R1 and R2 at the supports ?

a)77.5 N, 245.5 N

b)105.5 N 33.5 N

c)37.5 N, 312.5 N

d)27.5 N, 232.5 N

Answer 1

Answer on Question 59417, Physics, Mechanics, Relativity

Question:

A uniform plank AB $30m$ long, weighing $100N$ is pivoted at points P, Q which are $5m$ from the ends A and B respectively. A boy of weight $250N$ stands at point D on the plank, $1m$ away from $Q$ and the arrangement is in equilibrium. Determine the reaction $R_{1}$ and $R_{2}$ at the supports?

a) $R_{1} = 77.5N,R_{2} = 245.5N$

b) $R_{1} = 105.5N,R_{2} = 33.5N$

c) $R_{1} = 37.5N,R_{2} = 312.5N.$

d) $R_{1} = 27.5N,R_{2} = 232.5N.$

Solution:

a) To find the reaction force at the left support, $R_{1}$ , we should consider the sum of moments of forces around the point Q. From the condition of the question we know that the arrangement is in equilibrium, thus the sum of all moments is equal to zero:

$$\sum M _ {Q} = 0,$$ $$R _ {1} l _ {P Q} - W _ {p l a n k} l _ {C Q} + W _ {b o y} l _ {Q D} = 0.$$

From this equation, we can find $R_{1}$ :

$$R _ {1} = \frac {W _ {p l a n k} l _ {C Q} - W _ {b o y} l _ {Q D}}{l _ {P Q}} = \frac {1 0 0 N \cdot 1 0 m - 2 5 0 N \cdot 1 m}{2 0 m} = \frac {7 5 0 N \cdot m}{2 0 m} = 3 7. 5 N.$$

b) To find the reaction force at the right support, $R_{2}$, we should consider the sum of moments of forces around the point P. Again, since the arrangement is in equilibrium the sum of all moments is equal to zero:

$$\sum M_{P} = 0,$$ $$W_{plank} l_{PC} + W_{boy} l_{PD} - R_{2} l_{PQ} = 0.$$

From this equation, we can find $R_{2}$:

$$\begin{array}{l} R_{2} = \frac{W_{plank} l_{PC} + W_{boy} l_{PD}}{l_{PQ}} = \frac{100 N \cdot 10 m + 250 N \cdot 21 m}{20 m} = \frac{6250 N \cdot m}{20 m} = \\ = 312.5 N. \end{array}$$

Answer:

c) $R_{1} = 37.5 N, R_{2} = 312.5 N$.

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