
When forces ${F_1}$, ${F_2}$ and ${F_3}$ are acting on a particle of mass $m$ such that ${F_2}$ and ${F_3}$ are mutually perpendicular, then the particles remain stationary. If the force ${F_1}$ is now removed then the magnitude of the acceleration of the particle is:
Answer
568.8k+ views
Hint: As it is given in the question that particles are stationary, then the resultant of the forces will be zero. From the relation of resultant forces, we can find the value of force ${F_1}$. Now, for calculating the acceleration of the particle, we will use Newton's law of motion equation which is given below.
Formula used:
The equation of Newton’s second law of motion is given below
$F = ma$
$ \Rightarrow \,a = \dfrac{F}{m}$
Here, $F$ is the force acting on the particle, $m$ is the mass of the particle and $a$ is the acceleration of the particle.
Complete step by step answer:
Consider a particle of mass $m$ on which three forces ${F_1}$, ${F_2}$ and ${F_3}$ are acting. Now, when the particle remains stationary, then the resultant of the three forces will be zero and is given below
${F_1} + {F_2} + {F_3} = 0$
$ \Rightarrow \,{F_1} = - \left( {{F_2} + {F_3}} \right)$
Therefore, from the above relation, we can say that the magnitude of force ${F_1}$ is equal to the magnitude of the sum of forces ${F_2}$ and ${F_3}$ i.e. ${F_2} + {F_3}$, but the direction of these forces ${F_2} + {F_3}$ will be opposite to force ${F_1}$. Now, consider that the force ${F_1}$ is now removed from the particle of mass $m$, therefore, the magnitude of the force of particle acting on the mass $m$ will be
$\Rightarrow\,magnitude\,of\,\left( {{F_2} + {F_3}} \right)$
$\Rightarrow \, - \,magnitude\,of\,{F_1}$
Now, according to Newton’s second law of motion, the acceleration of the particle can be calculated as shown below
$F = ma$
$ \Rightarrow \,a = \dfrac{F}{m}$
Therefore, the acceleration of particle in case of magnitude of force when ${F_1}$ is removed is given below
$a = \dfrac{{{F_2} + {F_3}}}{m}$
$ \therefore \,a = - \dfrac{{{F_1}}}{m}$
Therefore, the magnitude of acceleration of the particle is $ - \dfrac{{{F_1}}}{m}$, but the direction of acceleration is opposite to the force ${F_1}$.
Note:Here, we got the value of acceleration as negative. The acceleration of the particle will be negative when the object moving will slow down. Also, we can say that the acceleration of the particle will be negative when the speed of the particle will decrease.
Formula used:
The equation of Newton’s second law of motion is given below
$F = ma$
$ \Rightarrow \,a = \dfrac{F}{m}$
Here, $F$ is the force acting on the particle, $m$ is the mass of the particle and $a$ is the acceleration of the particle.
Complete step by step answer:
Consider a particle of mass $m$ on which three forces ${F_1}$, ${F_2}$ and ${F_3}$ are acting. Now, when the particle remains stationary, then the resultant of the three forces will be zero and is given below
${F_1} + {F_2} + {F_3} = 0$
$ \Rightarrow \,{F_1} = - \left( {{F_2} + {F_3}} \right)$
Therefore, from the above relation, we can say that the magnitude of force ${F_1}$ is equal to the magnitude of the sum of forces ${F_2}$ and ${F_3}$ i.e. ${F_2} + {F_3}$, but the direction of these forces ${F_2} + {F_3}$ will be opposite to force ${F_1}$. Now, consider that the force ${F_1}$ is now removed from the particle of mass $m$, therefore, the magnitude of the force of particle acting on the mass $m$ will be
$\Rightarrow\,magnitude\,of\,\left( {{F_2} + {F_3}} \right)$
$\Rightarrow \, - \,magnitude\,of\,{F_1}$
Now, according to Newton’s second law of motion, the acceleration of the particle can be calculated as shown below
$F = ma$
$ \Rightarrow \,a = \dfrac{F}{m}$
Therefore, the acceleration of particle in case of magnitude of force when ${F_1}$ is removed is given below
$a = \dfrac{{{F_2} + {F_3}}}{m}$
$ \therefore \,a = - \dfrac{{{F_1}}}{m}$
Therefore, the magnitude of acceleration of the particle is $ - \dfrac{{{F_1}}}{m}$, but the direction of acceleration is opposite to the force ${F_1}$.
Note:Here, we got the value of acceleration as negative. The acceleration of the particle will be negative when the object moving will slow down. Also, we can say that the acceleration of the particle will be negative when the speed of the particle will decrease.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

Chemical formula of Bleaching powder is A Ca2OCl2 B class 11 chemistry CBSE

Name the part of the brain responsible for the precision class 11 biology CBSE

The growth of tendril in pea plants is due to AEffect class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

