Question

The resultant of the three concurrent space forces at A is 𝑹̅ = (-788𝒋̅) N. Find the magnitude of F1,F2 and F3 forces.

Given : A=(0,12,0)
            B=(-9,0,0)
           C=(0,0,5)
          D=(3,0,-4)
 Resultant of forces = (-788𝑗̅) N
To find : Magnitude of forces F1,F2,F3

Answer 1

Assume 𝑎̅,𝑏̅,𝑐̅ and 𝑑̅ be the position vectors of points A,B,C and D respectively w.r.t origin O
`bar(OA)` = 𝑎̅ = 12𝑗̅

`bar(OB)` = 𝑏̅ = -9𝑖̅
`bar(OC)`= 𝑐̅ = 5𝑘̅
`bar(OD)`= 𝑑̅ = 3𝑖̅ - 4𝑘̅

`bar(AB)`= 𝑏̅ - 𝑎̅
      = -9𝑖̅-12𝑗̅
`bar(AC) `= 𝑐̅ – 𝑎̅ = 5𝑘 ̅- 12𝑗̅
`bar(AD)`= 𝑑̅ - 𝑎̅ = 3𝑖̅ -12𝑗 ̅ - 4𝑘

Sr.no.	Vector	Magnitude
1.	`bar(AB)`	15
2.	`bar(AC) `	13
3.	`bar(AD)`	13

 

Sr.no.	Vector	`"Unit vector "= ("vector")/("Magnitude of vector")`
1.	`bar(AB)`	`(−3)/5 bar(i) - 4/5 bar(j)`
2.	`bar(AC) `	`(−12)/13 bar(j) - 5/13 bar(k)`
3.	`bar(AD)`	`(3)/13 bar(i) - 12/13 bar(j) - 4/13 bar(K)`

Force along `bar(AB) = bar(F1) = F1 ((−3)/5 bar(i) - 4/5 bar(j))`

Force along `bar(AC) = bar(F2) = F2((−12)/13 bar(j) - 5/13 bar(k))`

Force along  `bar(AB) = bar(F3) = F3( (3)/13 bar(i) - 12/13 bar(j) - 4/13 bar(K))`

Resultant force `bar(R) = bar(F1) +bar(F2)+ bar(F3)`

`-788 bar(j) = F1 ((−3)/5 bar(i) - 4/5 bar(j)+ bar(F2) (−12)/13 bar(j) - 5/13 bar(k) + bar(F3) (3)/13 bar(i) - 12/13 bar(j) - 4/13 bar(K))`

 

`0bar(j)-788 bar(j) + 0bar(k) = F1 ((−3)/5 bar(i) - 4/5 bar(j)+ bar(F2) (−12)/13 bar(j) - 5/13 bar(k) + bar(F3) (3)/13 bar(i) - 12/13 bar(j) - 4/13 bar(K))`

Comparing the equation on both sides 

`(−3F1)/5 + (3F3)/13=0 `………..(1)
`-(4F1)/5--(12F2)/13 -(12F2)/13=-788` ………(2)

`(5F2)/13 -(4F3)/13 = 0 ` ……………(3)
Solving (1),(2) and (3)
F1=153.9063 N
F2=320.125 N
F3=400.1563 N

Sr.no.	Force	Magnitude
1.	F1	153.9063 N
2.	F2	320.125 N
3.	F3	400.1563 N