coefficient of friction?

Question

The coefficient of static friction between a box and an inclined plane is 0.35. What is the minimum angle required for the box to begin sliding down the incline?

Answer 1

The frictional force is coefficient of friction times the normal reaction.



If ‘m’ is the mass of the object and ‘a’ is the angle of the plane, then the normal reaction is mg cos a.



The force that is pulling the object along the plane downward is mg sin a.



This is equal to the frictional force.



Therefore, mg sin a = (mu) mg cos a;



(mu) is the coefficient of friction.



Hence (mu) = tan a. We need not know the weight of the object.



This we can have as a formula for these cases of inclined plane.



Given 0.35 = tan a;



Hence a = 19.3 degree.





Tan a is opposite side / adjacent side of the inclined plane.



As long as "a" is constant the ratio will not change whatever is the length and height.



Use a calculator to find the tan inverse of 0.35.



Or check the value of tan 19.3 . you will get 0.35.



Contact me if you have any doubt to my email address given in my profile. [e-mail].

Answer 2

On the inclined plane you split the gravitational force upon the box on two directions, Gx on the direction of the inclined plane and Gy on the direction normal to the inclined plane.

From geometry results that Gx = G*sin a and Gy = G*cos a, where G is the original grav. of the box and a is the required angle.

If the box moves without acceleration, that means that Gx is equal to the friction force opposing the movement, which is F=u*N, with u - that coefficient that is given and N - the normal force (sorry, I know the names in my own language, maybe the translation is not accurate) which in our case is exactly Gy.

So, Gx=F

G*sina=u*G*cosa results u=(G*sina)/(G*cosa), you simplify and

u = tangent a or a= arc tangent u

Answer 3

the angle of inclination for the inclined plan required to just start sliding motion by the box is 19.29 degree. (tan inverse 0.35)

the formula is

coefficient of static friction = tan ( angle of inclination of the plane)

Answer 4

The gravitational force can be split into a component along the plane, which is proportional to the since of the angle, and a component perpendicular to the plane, which is proportional to the cosine of the angle.



The frictional force is 0.35 times the latter. So you must solve the equation

0.35 cos alpha = sin alpha.

Answer 5

Depends on the tempature if there is ice on the incline the slide will occur at about 18.259741degrees, however if the incline is located in southern California the incline will increase to 34.498727 degrees.

Answer 6

Diagram the forces on the block. chop up gravity into aspects alongside the incline and perpendicular to it. stability the forces to place in writing an expression for the finished tension in terms of the coefficient of kinetic friction. positioned that into the equation F = ma. sparkling up for ?.

Answer 7

need to know the weight of the box