
A monkey is sitting on a tree limb. The limb exerts a normal force of \[48N\] and a frictional force of \[20N\]. Find the magnitude of the total force exerted by the limb on the monkey.
\[(A)23N\]
\[(B)36N\]
\[(C)\;48N\]
\[(D)52N\;\]
Answer
498.3k+ views
Hint: Once we have two bodies that are in contact with one another, then there acts a force in between them. Since this force is acting because of the contact between the two bodies, it is termed as the contact force. This contact force may be resolved in two components. The vertical component of the contact force is understood because the normal reaction force and also the horizontal component is called the frictional force. The normal reaction force is that which acts perpendicular to the surfaces in contact. And also the frictional force is that which resists the motion between two surfaces.
Formula used:
The resultant magnitude of the force \[R = \sqrt {{{\left( {{F_n}} \right)}^2} + {{\left( {{F_f}} \right)}^2}} \]
Where, \[{F_n}\]-Normal force and \[{F_f}\]-Frictional force
Complete step-by-step solution:
The limb exerts a normal force of 48N and a frictional force of 20N then the magnitude of the entire force exerted by the limb on the monkey is
\[R = \sqrt {{{\left( {{F_n}} \right)}^2} + {{\left( {{F_f}} \right)}^2}} \]
\[ \Rightarrow R = \sqrt {{{\left( {48} \right)}^2} + {{\left( {20} \right)}^2}} \]
\[ \Rightarrow R = \sqrt {2304 + 400} = \sqrt {2704} \]
\[ \Rightarrow R = 52N\]
The magnitude of the total force exerted by the limb on the monkey is\[52N\].
Hence, option D is correct.
Additional information:
The coefficient of friction (COF) may be a dimensionless number and COF is termed as the ratio between friction force and normal force. The material which has COF smaller than \[0.1\] are considered lubricous materials. COF depends on the character of the materials and surface roughness.
Note:When two bodies are placed in grips, attractive forces act between their particles at the surface of contact. As a result, each body exerts a contact force on another body. These mutual forces are equal and opposite by obeying Newton’s third law. A component of the contact force is termed the frictional force.
Formula used:
The resultant magnitude of the force \[R = \sqrt {{{\left( {{F_n}} \right)}^2} + {{\left( {{F_f}} \right)}^2}} \]
Where, \[{F_n}\]-Normal force and \[{F_f}\]-Frictional force
Complete step-by-step solution:
The limb exerts a normal force of 48N and a frictional force of 20N then the magnitude of the entire force exerted by the limb on the monkey is
\[R = \sqrt {{{\left( {{F_n}} \right)}^2} + {{\left( {{F_f}} \right)}^2}} \]
\[ \Rightarrow R = \sqrt {{{\left( {48} \right)}^2} + {{\left( {20} \right)}^2}} \]
\[ \Rightarrow R = \sqrt {2304 + 400} = \sqrt {2704} \]
\[ \Rightarrow R = 52N\]
The magnitude of the total force exerted by the limb on the monkey is\[52N\].
Hence, option D is correct.
Additional information:
The coefficient of friction (COF) may be a dimensionless number and COF is termed as the ratio between friction force and normal force. The material which has COF smaller than \[0.1\] are considered lubricous materials. COF depends on the character of the materials and surface roughness.
Note:When two bodies are placed in grips, attractive forces act between their particles at the surface of contact. As a result, each body exerts a contact force on another body. These mutual forces are equal and opposite by obeying Newton’s third law. A component of the contact force is termed the frictional force.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

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

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

