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
1.1k views
in Laws of motion by (52.4k points)

The 1350-kg car is at rest on a plane surface with its brakes locked. The unit vector en = 0.231 i + 0.923 j +0.308 k is perpendicular to the surface. The y axis points upward. The direction cosines of the cable from A to B are cos θx = -0.816, cos θy = 0.408, cos θz = -0.408, and the tension in the cable is 1.2 kN. Determine the magnitudes of the normal and friction forces the car’s wheels exert on the surface.

1 Answer

+1 vote
by (48.4k points)
selected by
 
Best answer

Assume that all forces act at the center of mass of the car. The vector equation of equilibrium for the car is

FS + TAB + W = 0.

Writing these forces in terms of components, we have

Substituting these values into the equations of equilibrium and solving for the unknown components of FS, we get three scalar equations of equilibrium. These are:

Substituting in the numbers and solving, we get

The next step is to find the component of FS normal to the surface. This component is given by

FN = FN x en = FSxeny + FSxeny + FSzenz.

Substitution yields

FN = 12149 N.

From its components, the magnitude of FS is FS = 12800 N. Using the Pythagorean theorem, the friction force is

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.

...