JEE Main 202225 JulyEvening Shift
- Exam
- JEE Main
- Year
- 2022
- Date
- 25 July
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Laws of Motion
JEE Main 2022 · 25 July · Evening Shift · Physics · Laws of Motion
For a free body diagram shown in the figure, the four forces are...
For a free body diagram shown in the figure, the four forces are applied in the '' and '' directions. What additional force must be applied and at what angle with positive -axis so that the net acceleration of body is zero?
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Let additional force required be .
Vector sum of the four forces and additional force.
\overset{\to}{F} + 5\overset{^}{i} - 6\overset{^}{i} + 7\overset{^}{j} - 8\overset{^}{j} = 0
Force required to make net acceleration zero
\overset{\to}{F} = \overset{^}{i} + \overset{^}{j}
Therefore, the required force is \left|\overset{\to}{F}\right| = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}Nxtan\theta = \dfrac{ycomponent}{xcomponent} = \dfrac{1}{1}\theta = 45 °