Which process involves transformation of liquid water
2 answers:
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Answer:
Komila should sit 0.33m from the middle of the board towards tahreen.
Explanation:
We are told to treat each student as point-like objects. So i have attached a rigid body diagram to depict this.
From the diagram,
F_d is force exerted by dan
F_t is force exerted by tahreen
F_k is force exerted by komila
F_b is force of board at the mid point.
x1 is distance of dan from the centre of the chair
x2 is distance of komila from the centre of the chair
x3 is distance of tahreen from komila
We are given;
Mass of Dan;m_d = 62 kg
Mass of tahreen;m_t = 50 kg
Mass of komila;m_k = 54 kg
Now, taking moments about the centre of the chair, we have;
(F_d*x1) - (F_k*x2) - (F_t(x2 + x3)) = 0
Now,F_d = m_d*g ; F_t = m_t*g ; F_k = m_k*g
We are told that the board is 3m long. So, if we assume that the fulcrum position of the chair coincides with the midpoint of boards length, we'll have;
x1 = (x2 + x3) = 1.5
Thus, we now have;
(F_d*1.5) - (F_k*x2) - (F_t*1.5) = 0
F_d = m_d*g = 62 * 9.8 = 607.6 N
F_t = m_t * g = 50 x 9.8 = 490 N
F_k = m_k * g = 54 x 9.8 = 529.2 N
So plugging in these values, we have;
(607.6 * 1.5) - (529.2 * x2) - (490 * 1.5) = 0
911.4 - 735 = 529.2 x2
529.2 x2 = 176.4
x2 = 176.4/529.2
x2 = 0.33m
Komila should sit 0.24m from the middle of the board towards tahreen
Answer:
B
Explanation:
Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.
