Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Three forces <img alt="F_1=10 \mathrm{~N}, F_2=8 \mathrm{~N}, \mathrm{~F}_3=6 \mathrm{~N}" src="https://entrancecorner.oncodecogs.com/gif.latex?F_1%3D10%20%5Cmathrm%7B%7EN%7D%2C%20F_2%3D8%20%5Cmath

Three forces F_1=10 \mathrm{~N}, F_2=8 \mathrm{~N}, \mathrm{~F}_3=6 \mathrm{~N} are acting on a particle of mass 5 \mathrm{~kg}. The forces F_2 and F_3 are applied perpendicularly so that particle remains at rest. If the force \mathrm{F}_1 is removed, then the acceleration of the particle is :

Option: 1

4.8 \mathrm{~ms}^{-2}


Option: 2

0.5 \mathrm{~ms}^{-2}


Option: 3

7 \mathrm{~ms}^{-2}


Option: 4

2 \mathrm{~ms}^{-2}


Answers (1)

best_answer

\begin{aligned} F_1 & =10 \mathrm{~N} \\ \mathrm{~F}_2 & =8 \mathrm{~N} \\ \mathrm{~F}_3 & =6 \mathrm{~N} \\ \overrightarrow{\mathrm{a}} & =\frac{\mathrm{F}_2 \hat{\mathrm{i}}+\mathrm{F}_3 \hat{\mathrm{j}}}{5} \\ \overrightarrow{\mathrm{a}} & =\frac{8 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}}{5} \\ |\overrightarrow{\mathrm{a}}| & =\frac{\sqrt{64+36}}{5} \\ & =\frac{10}{5}=2 \mathrm{~m} / \mathrm{sec}^2 \end{aligned}

OR

\mathrm{F}_1^2=\mathrm{F}_2^2+\mathrm{F}_3^2

Now F1 is removed then net force

\begin{aligned} & \mathrm{F}_{\text {net }}=\sqrt{\mathrm{F}_2^2+\mathrm{F}_3^2} \\ & =\sqrt{64+36} \\ & \mathrm{~F}_{\text {net }}=10 \\ & 10=5 \mathrm{a} \\ & \mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^2 \end{aligned}

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question