Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.8k views
in Laws of motion by (52.4k points)

The 14000 N car is at rest on the plane surface. The unit vector en = 0.456i + 0.570j + 0.684k is perpendicular to the surface. Determine the magnitudes of the total normal force N and the total friction force f exerted on the surface by the car’s wheels.

1 Answer

+1 vote
by (48.5k points)
selected by
 
Best answer

The forces on the car are its weight, the normal force, and the friction force.

The normal force is in the direction of the unit vector, so it can be written

N = Nen = N(0.456i + 0.570j + 0.684k)

The equilibrium equation is 

Nen + f - (14000 N)j =0

The friction force f is perpendicular to N, so we can eliminate the friction force from the equilibrium equation by taking the dot product of the equation with en. 

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...