
A uniform rod of mass $m = 15kg$ leans against a smooth vertical wall making an angle $\theta = {37^ \circ }$ with horizontal. The other end rests on a rough horizontal floor. Calculate the normal force and the friction force that the floor exerts on the rod.
Answer
160.8k+ views
Hint: In this problem we are going to write first all the forces which are acting on the rod. These include gravitational force, normal force and frictional force. And then by applying force balance we are going to get normal and friction force that the floor exerts on the rod.
Complete answer:
The forces acting on the rod are
(a) Its weight W,
(b) normal force N, by the vertical wall,
(c) normal force $N_2$ by the floor and
(d) frictional force f by the floor.
Taking horizontal and vertical components,
${N_1} = f$------(1)
And ${N_2} = mg$------(2)
Taking torque about ground we get,
${N_1}x\cos \theta = mg\dfrac{x}{2}\sin \theta $
$ \Rightarrow {N_1}\dfrac{3}{5} = mg\dfrac{1}{2} \times \dfrac{4}{5}$
$ \Rightarrow {N_1} = mg \times \dfrac{2}{3}$
So, ${N_1} = 100N$ also the$f = 100N$
The normal force by floor ${N_2} = mg = 150N$
Note: While attempting this type of question, always remember to apply force balance. The balance forces plays an important role to solve the above question. It is important note that, in this case we considered the object to be at static motion thus we haven’t used any value related to motion and velocity.
Complete answer:
The forces acting on the rod are
(a) Its weight W,
(b) normal force N, by the vertical wall,
(c) normal force $N_2$ by the floor and
(d) frictional force f by the floor.
Taking horizontal and vertical components,
${N_1} = f$------(1)
And ${N_2} = mg$------(2)
Taking torque about ground we get,
${N_1}x\cos \theta = mg\dfrac{x}{2}\sin \theta $
$ \Rightarrow {N_1}\dfrac{3}{5} = mg\dfrac{1}{2} \times \dfrac{4}{5}$
$ \Rightarrow {N_1} = mg \times \dfrac{2}{3}$
So, ${N_1} = 100N$ also the$f = 100N$
The normal force by floor ${N_2} = mg = 150N$
Note: While attempting this type of question, always remember to apply force balance. The balance forces plays an important role to solve the above question. It is important note that, in this case we considered the object to be at static motion thus we haven’t used any value related to motion and velocity.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
