The forces 𝐹 one, which is equal to two 𝑖 plus seven 𝑗; 𝐹 two, which is equal to 𝑎𝑖 minus six 𝑗; and 𝐹 three, which is equal to six 𝑖 plus 𝑏 plus eight 𝑗, act on a particle, where 𝑖 and 𝑗 are two perpendicular unit vectors. Given that the system is in equilibrium, determine the values of 𝑎 and 𝑏.
If the system is in equilibrium, then the resultant force must equal zero. This means that the 𝑖-components — two 𝑖, 𝑎𝑖, and six 𝑖 — must equal zero. The coefficients are two, 𝑎, and six. Therefore, two plus 𝑎 plus six equals zero. Two plus six is equal to eight. Therefore, 𝑎 plus eight is equal to zero. Subtracting eight from both sides of the equation gives us a value of 𝑎 of negative eight. This means that the force 𝐹 two is negative eight 𝑖 minus six 𝑗.
As the 𝑗-components must also equal zero, seven 𝑗 minus six 𝑗 and 𝑏 plus eight 𝑗 must equal zero. Once again, the coefficients are seven, negative six, and 𝑏 plus eight. Seven minus six plus 𝑏 plus eight equals zero. Seven take away six is one. One plus eight is equal to nine. Therefore, 𝑏 plus nine equals zero. Subtracting nine from both sides of this equation gives us a value of 𝑏 equal to negative nine.
This means that the force 𝐹 three is equal to six 𝑖 plus negative nine plus eight 𝑗. Negative nine plus eight is negative one. Therefore, 𝐹 three is six 𝑖 minus one 𝑗. If the three forces 𝐹 one, 𝐹 two, and 𝐹 three are acting on a particle where the system is in equilibrium, then the value of 𝑎 is negative eight and the value of 𝑏 is negative nine.