Name: Two Block System
Uploaded: 2019-07-19
Duration: 6 min 20 s
Description: Two blocks placed on a surface exerts a force between them and also with the surface. Let us discuss how these forces can be solved numerically in this video.

Question

What is the minimum value of force (in the following two cases) required to pull a block of mass

m

on a horizontal surface having a coefficient of friction

μ

? Also find the angle this force makes with the horizontal.
(a) If force is parallel to a horizontal surface.
(b) If force is in any direction (Also find the angle this force makes with the horizontal.)

Accepted Answer

$(a)$ Consider the FBD shown in the below figure.

Here, the normal reaction should be equal to the weight of the block.

Then the limiting friction force $(f_{L})$ on the block is equal to minimum force required to pull a block.

Equations from FBD,

$F = f_{L}$

Since, $f_{L} = μ N = μ M g$

Therefore, $F = μ M g$

$(b)$ Consider the situation shown in the figure below. Applied a force $F$ at $θ$ angle from horizontal.

So, the force required to pull the body is,

$F \cos θ = f$

$\Rightarrow F \cos θ = μ N (N = μ M g - F \sin θ)$

$\Rightarrow F \cos θ = μ M g - μ F \sin θ$

$\Rightarrow F \cos θ + μ F \sin θ = μ M g$

$\Rightarrow F \cos θ + μ \sin θ = μ M g$

$\Rightarrow F = \frac{μ M g}{\cos θ + μ \sin θ}$

For $F_{\min} \Leftrightarrow \cos θ + μ \sin θ \to \max$

$\frac{d}{d θ} [\cos θ + μ \sin θ] = 0$

$\Rightarrow - \sin θ + μ \cos θ = 0$

$\Rightarrow μ = \tan θ = \frac{sinθ}{cosθ}$

So, $\Rightarrow F_{\min} = \frac{μ M g}{\cos θ + μ \sin θ}$

$= \frac{μ M g}{\cos θ + (\frac{\sin θ}{\cos θ}) \sin θ}$

$\Rightarrow F_{\min} = \frac{μ M g \cos θ}{\cos^{2} θ + \sin^{2} θ}$

$\Rightarrow F_{\min} = μ M g \cos θ$

Important Questions on Friction

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