Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • For a free body diagram shown in the figure, the four forces are applied in the <img alt="\mathrm{ ' x ' \, and \: ' y ' }" src="/latex-image/?%5Cmathrm%7B%20%27%20x%20%27%20%5C%2C%20and%20%5C%3A%2

For a free body diagram shown in the figure, the four forces are applied in the \mathrm{ ' x ' \, and \: ' y ' }directions. What additional force must be applied and at what angle with positive \mathrm{\mathrm{x}-axis} so that the net acceleration of body is zero ?



 

Option: 1

\mathrm{\sqrt{2}N,45^{0}}


Option: 2

\mathrm{\sqrt{2} N, 135^{\circ}}
 


Option: 3

\mathrm{\frac{2}{\sqrt{3}} N, 30^{\circ}}


Option: 4

\mathrm{2 N, 45^{\circ}}


Answers (1)

best_answer

For net acceleration of body to be zero,

\mathrm{\sum \bar{F}=0}

\mathrm{ i.e., \sum F_X=0\: and\: \sum F_Y=0}

\mathrm{\sum F_x=(+5)+(-6)+F \cos \theta=0}

\mathrm{F \cos \theta=1 \rightarrow (1)}

\mathrm{\sum F_y=+7+(-8)+F \sin \theta=0}
                    \mathrm{F \sin \theta=1 \rightarrow (2)}

\mathrm{(1) \div (2)}

\mathrm{\tan \theta =1}

\mathrm{\theta=45^{\circ}}

\mathrm{F=\frac{1}{\cos \theta}=\sqrt{2}}

Hence (1) is correct option

Posted by

Ajit Kumar Dubey

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

Andhra launches free JEE, NEET coaching to over 1 lakh govt school students
anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions
anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel

Latest Question