A mass of 3kg rests on a rough plane inclined at 60 degrees to the horizontal. The coefficient of friction is 1/5. Find the force P acting parallel to the plane applied to the mass, in order to just prevent motion down the plane.

Firstly draw a diagram of the problem.Then resolve the forces into their components parallel and perpendicular to the plane.Resolving parallel: P + Fmax = 3gsin(60) equation 1.Resolving perpendicular: R = 3gcos(60) =14.7N equation 2. Then substitute Fmax= Mu.R into equation 1 and then sub R (equation 2) into equation 1. Then solve to find P. P = 3gsin(60)- Mu.R P = 3gsin(60) - 0.2x14.7= 22.5 N

JM
Answered by Jonathan M. Maths tutor

4248 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Express (1 + 4 * 7^0.5)/(5 + 2 * 7^0.5) in the form m + n * 7^0.5


If the functions f and g are defined: f: x--> x/5 + 4 g : x--> 30x + 10. what is x, if fg(x) = x. ?? What would fgf(x) = x^2 be??


What is the velocity of the line from vector A(3i+2j+5k) to vector B(10i-3j+2k)?


Solve D/dx (ln ( 1/cos(x) + tan (x) )


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences