a particle acted upon by the forces vector F1=2i+aj-3k vector F2=5i+cj-bk vector F3=bi+5j-7k vector F4=ci+6j-ak.find out the values of the constants a, b, c, in order that particle will be in equilibrium.
Answers
Answer:
Explanation:
=> Here, A particle acted upon by the forces vector :
F1=2i+aj-3k
F2=5i+cj-bk
F3=bi+5j-7k
F4=ci+6j-ak
=> when the sum of all forces will be zero vector, the particle will be in equilibrium.
Thus, F1 + F2 + F3 + F4 = 0
(2i+aj-3k) + (5i+cj-bk) + (bi+5j-7k) + (ci+6j-ak) = Ray 0
(2+5+b+c)i + (a+c+5+6)j + (-3-b-7-a)k = Ray 0
(7+b+c)i + (a+c+11)j + (10+b+a)k = 0i +0j + 0k
[Because, ray 0 = 0i + 0j + 0k ]
So,
7 + b + c = 0 ...(i)
11 + a + c = 0 ...(ii)
10 + a + b = 0 ...(iii)
=> By adding (i), (ii) and (iii), we get
28 + 2(a+b+c) = 0 or a + b + c + 14 = 0 ...(iv)
=> If we subtract eq (i) from eq (iv), we get
(a+b+c+14) - (7+b+c) = 0
a + 7 = 0
a = -7
=> By placing the value of a in eq (ii), we get
11 + (-7) + c = 0
c = -4
=> By placing the value of c in eq (i), we obtain
7 + b + (-4) = 0
b = -3
Thus, the value of the constants a, b, c are -7, -3 and -4 respectively.
Answer:
Explanation:
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