Answer on Question 58849, Physics, Mechanics, Relativity
Question:
13. 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 R1 and R2 at the supports.
a) R1=77.5N,R2=245.5N
b) R1=105.5N,R2=33.5N
c) R1=37.5N,R2=312.5N.
d) R1=27.5N,R2=232.5N.
Solution:

a) To find the reaction force at the left support, R1 , 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:
∑MQ=0,R1lPQ−WplanklCQ+WboylQD=0.
From this equation, we can find R1 :
R1=lPQWplanklCQ−WboylQD=20m100N⋅10m−250N⋅1m=20m750N⋅m=37.5N.
b) To find the reaction force at the right support, R2, 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:
∑MP=0,WplanklPC+WboylPD−R2lPQ=0.
From this equation, we can find R2:
R2=lPQWplanklPC+WboylPD=20m100N⋅10m+250N⋅21m=20m6250N⋅m==312.5N.
Answer:
c) R1=37.5N,R2=312.5N.
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