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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 4.0m4.0\,m just begins to slide down when the upper end of the plane is 1.0m1.0\,m high from the ground. Calculate the coefficient of static friction.

Explanation

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=FFN{\mu _s} = \dfrac{{\vec F}}{{{{\vec F}_N}}}.

Complete step by step answer:
Let us first draw the diagram, Draw a line ABAB representing inclined plane of length 4m4m and a perpendicular height CBCB of length 1m1m and form a right angle triangleABCABC.

Here, in the diagram OO represents the object and mgmg is the force of gravity acting on the object. Let θ\theta be the angle between ACAC and ABAB then, from the geometry of figure, the applied force acting on the object is represented by F\vec F and the normal reaction on object is represented by F2{\vec F_2}.From geometry, we can see that the magnitude of force F\vec F is given by,
F=mgsinθ(i)\vec F = mg\sin \theta \to (i)

And the magnitude of normal reaction force F2=mgcosθ(ii){\vec F_2} = mg\cos \theta \to (ii)
We also know that coefficient of static friction μs=FFN{\mu _s} = \dfrac{{\vec F}}{{{{\vec F}_N}}}
Using equation (i)(i) and equation (ii)(ii) we have,
μs=mgsinθmgcosθ μs=tanθ(iii) {\mu _s} = \dfrac{{mg\sin \theta }}{{mg\cos \theta }} \\\ \Rightarrow {\mu _s} = \tan \theta \to (iii) \\\
Now In right angle triangleABCABC, using Pythagoras theorem
AB=161AB = \sqrt {16 - 1}
AB=15\Rightarrow AB = \sqrt {15}
Again, in triangle ABCABC
tanθ=BCAB tanθ=115 tanθ=0.258 \tan \theta = \dfrac{{BC}}{{AB}} \\\ \Rightarrow \tan \theta = \dfrac{1}{{\sqrt {15} }} \\\ \Rightarrow \tan \theta = 0.258 \\\
From equation (iii)(iii) we get
μs=0.258\therefore {\mu _s} = 0.258

Hence, the coefficient of static friction is μs=0.258{\mu _s} = 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.