Solving Friction Problems

Solving Friction Problems-74
Atantanθ Solution 3 The downward force=mgsinθ The upward force=μmgcosθ Net force f(y)=mgsinθ-μmgcosθ =mg(sinθ-kycosθ) at y=A, f(y)=0 0=sinθ-k Acosθ or K=tanθ/A Question 4A given object takes n times as much times to slide down a 45 rough incline as its takes to slide down a perfectly smooth 45 incline. μL/(1-μ) Solution 5 Let m be the mass per unit length Let L be the full length and l be the length of chain hanging So net force downwards=mlg Net frictional force in opposite direction=μm(L-l)g Now mlg=μm(L-l)g or l=μ(L-l) or l=μL/(1 μ) Question 6The coefficient of static and kinetic friction between a body and the surface are .75 and .50 respectively.

Atantanθ Solution 3 The downward force=mgsinθ The upward force=μmgcosθ Net force f(y)=mgsinθ-μmgcosθ =mg(sinθ-kycosθ) at y=A, f(y)=0 0=sinθ-k Acosθ or K=tanθ/A Question 4A given object takes n times as much times to slide down a 45 rough incline as its takes to slide down a perfectly smooth 45 incline. μL/(1-μ) Solution 5 Let m be the mass per unit length Let L be the full length and l be the length of chain hanging So net force downwards=mlg Net frictional force in opposite direction=μm(L-l)g Now mlg=μm(L-l)g or l=μ(L-l) or l=μL/(1 μ) Question 6The coefficient of static and kinetic friction between a body and the surface are .75 and .50 respectively.

Tags: Controversial EssaysBusiness Planning CoursesPaper Chromatography TermsJoan Didion Online EssaysReflective Journal Summary EssayRole Of Teachers In Character Building EssayPurpose Of Writing A Reflective EssayComputer-Assisted Language Learning ThesisThesis And Argumentative EssayEssay About Speech Choir

/2μg from v=u at or t=v/μg Question 2A horizontal force of F N is necessary to just hold a block stationary against a wall. F/μ d none of these Solution 2 Let W be the weight Reaction force=F Weight downward=W weight Upward=frictional force=μr=μF For no movement weight Upward=Weight downward W=μF Question 3. If the coefficient of friction between the chain and the table top is μ. What is the shortest distance and shortest time in which the block can be stopped if μ is coefficient of friction a.v/2μg from v=u at or t=v/μg Question 8A horizontal force of F N is necessary to just hold a block stationary against a wall. F/μ d none of these Solution 8 Let W be the weight Reaction force=F Weight downward=W weight Upward=frictional force=μr=μF For no movement weight Upward=Weight downward W=μF Note to our visitors :- Thanks for visiting our website.

The coefficient of friction between the block and the wall is μ. A body is sliding down a rough inclined plane of angle of inclination θ for which coefficient of friction varies with distance y as μ(y)=Ky where K is constant. what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on table.? The coefficient of friction between the block and the wall is μ. DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU.

You can use this tool as a coefficient of friction calculator, too.

Static friction acts when the object remains stationary. If we don't take friction into account, even the smallest force should cause some acceleration of the box according to Newton's second law.

Do notice that there is a forward force on the lower carton in addition to the frictional force acting backwards on it, and so you do need to justify to yourself that it doesn’t affect that answer.

Regardless, this problem is provided as a sample because it illustrates how you would apply Newton’s 2nd and 3rd Law in a wide variety of problems that you might be asked to solve on your homework, and so it is worth looking at the mathematical analysis. At the start of the problem, the truck and both cartons are traveling forward.

It is free, awesome and will keep people coming back!

A delivery truck with two stacked cartons in the back brakes quickly.

In reality, you need to pull quite hard for the box to start moving because of the static friction force.

Kinetic friction acts on a moving object or, in other words, on an object with nonzero kinetic energy.

SHOW COMMENTS

Comments Solving Friction Problems

The Latest from book-old2.ru ©