Will a 5 kg Block Slide on a 50 Degree Incline Plane? Calculation and Answer

[tony873004]

This is just a question I made up because there'll probably be one like this on our final exam.

A 5 kg block is on an incline plane of 50 degrees. coeffecient of friction is 0.5. Will it slide?

N = mg cos(50)
N = 31.529
friction = mu * N
friction = 15.764

force down = ma

f =m g cos(50)
f = 31.529

force down > friction so it will slide.

 

[Pyrrhus]

Actually the force making it slide it's the x component of gravity [itex] mg \sin \theta [/itex]

and yes if [itex] mg \sin \theta > \mu mg \cos \theta [/itex] then it will slide.

 

[tony873004]

Actually the force making it slide it's the x component of gravity [itex] mg \sin \theta [/itex]

and yes if [itex] mg \sin \theta > \mu mg \cos \theta [/itex] then it will slide.

Thank you...

 

[Pyrrhus]

Why don't you try finding the coefficient of friction that will avoid the system from sliding or finding the angle it will not slide an a specific coefficient of friction? Those are some of the common problems asked.

 

[tony873004]

Why don't you try finding the coefficient of friction that will avoid the system from sliding or finding the angle it will not slide an a specific coefficient of friction? Those are some of the common problems asked.

if it slides when mg sin theta > mg cos theta, then when they equal that should be the minimum mu necessary to keep it from sliding.
mg sin theta = mu mg cos theta
mu = (mg sin theta) / (mg cos theta)

but this gives me an answer greater than 1.

 

[Pyrrhus]

Because the inequality says [itex] \tan \theta > \mu [/tex] that's why you are getting a greater value.