Question
Question: A block placed on a rough inclined plane of length \(4.0\,m\) just begins to slide down when the upp...
A block placed on a rough inclined plane of length 4.0m just begins to slide down when the upper end of the plane is 1.0m high from the ground. Calculate the coefficient of static friction.
Solution
In static friction, the frictional force resists force that is applied to an object, and object remains at rest until force of static friction is overcome. Mathematically the coefficient of static friction is defined as ratio of applied force to the normal reaction. It’s denoted by μs=FNF.
Complete step by step answer:
Let us first draw the diagram, Draw a line AB representing inclined plane of length 4m and a perpendicular height CB of length 1m and form a right angle triangleABC.
Here, in the diagram O represents the object and mg is the force of gravity acting on the object. Let θ be the angle between AC and AB then, from the geometry of figure, the applied force acting on the object is represented by F and the normal reaction on object is represented by F2.From geometry, we can see that the magnitude of force F is given by,
F=mgsinθ→(i)
And the magnitude of normal reaction force F2=mgcosθ→(ii)
We also know that coefficient of static friction μs=FNF
Using equation (i) and equation (ii) we have,
μs=mgcosθmgsinθ ⇒μs=tanθ→(iii)
Now In right angle triangleABC, using Pythagoras theorem
AB=16−1
⇒AB=15
Again, in triangle ABC
tanθ=ABBC ⇒tanθ=151 ⇒tanθ=0.258
From equation (iii) we get
∴μs=0.258
Hence, the coefficient of static friction is μs=0.258.
Note: Normal reaction force is a contact force. It’s perpendicular to the surface and applied force is always in the direction of motion of an object. Coefficient of static friction is a dimensionless constant that characterizes the nature of the contact between the two surfaces.