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The angle of friction for two surfaces in contact is defined as the angle that the maximum contact force makes with the direction of normal force as shown in the figure. From the figure,
$$\begin{array}{l}\mathrm{tan}\alpha =\frac{{F}_{s}}{N}={\mu}_{s}\end{array}$$
The angle of friction, therefore, is :
$$\begin{array}{l}\Rightarrow \alpha ={\mathrm{tan}}^{-1}\left(\frac{{F}_{s}}{N}\right)={\mathrm{tan}}^{-1}\left({\mu}_{s}\right)\end{array}$$
We must understand here that angle of friction is a function of the coefficient of friction between two surfaces. This means that it depends on the nature of surfaces and is unique for the two surfaces in contact. Specially, we should note that it does not depend on maximum static force or normal force. If we increase “N” (say by putting more weight on the block), then maximum static friction also increases as ${F}_{s}={\mu}_{s}N$ and the ratio “ $\frac{{F}_{s}}{N}$ ” remains constant.
External force is required to initiate motion against friction force. For every situation, there is a particular direction in which the requirement of external force to initiate motion is a minimum. Here we consider the case of a block placed on a horizontal surface, which is pulled up by an external force, F, as shown in the figure.
At first consideration, we may reason that force would be always least when it is applied exactly opposite to the friction. The force required in such case is given by :
$$\begin{array}{l}{F}_{||}={\mu}_{s}N={\mu}_{s}mg\end{array}$$
This consideration, however, is not correct. The external force, besides overcoming maximum static friction, also affects normal force. The component of force in the vertical direction decreases normal force. Since friction is proportional to normal force, an unique orientation will require minimum magnitude of external force.
When angle “θ” increases (cosθ decreases), the force overcoming maximum static friction decreases :
$$\begin{array}{l}{F}_{||}={F}_{x}=F\mathrm{cos}\theta \end{array}$$
On the other hand, normal force, N, decreases with increasing angle (increasing sinθ),
$$\begin{array}{l}N=mg-F\mathrm{sin}\theta \end{array}$$
Consequently, the maximum static friction decreases as :
$$\begin{array}{l}{F}_{s}={\mu}_{s}N={\mu}_{s}(mg-F\mathrm{sin}\theta )\end{array}$$
Thus, there is a particular value of the angle “θ” for which magnitude of external force is minimum. We shall work out the particular orientation for this arrangement in the example given here.
Problem : An external force, "F", is applied on a block at an angle "θ" as shown in the figure. In order to initiate motion, what should be the angle at which magnitude of external force, F, is minimum? Also find the magnitude of minimum force.
Solution : We should obtain an expression of external force as a function of “θ” and then evaluate the function for obtaining the condition for the minimum value. We draw the free body diagram superimposed on the body diagram for the condition of maximum static friction as shown in the figure.
$$\begin{array}{l}\sum {F}_{x}=F\mathrm{cos}\theta -{\mu}_{s}N=0\\ \Rightarrow F\mathrm{cos}\theta ={\mu}_{s}N\end{array}$$
and
$$\begin{array}{l}\sum {F}_{y}=N+F\mathrm{sin}\theta -mg=0\\ \Rightarrow N=mg-F\mathrm{sin}\theta \end{array}$$
Combining two equations, we have :
$$\begin{array}{l}\Rightarrow F=\frac{{\mu}_{s}mg}{\mathrm{cos}\theta +{\mu}_{s}\mathrm{sin}\theta}\end{array}$$
But, we know that ${\mu}_{s}$ = tan α, where “α” is the angle of friction.
$$\begin{array}{l}\Rightarrow F=\frac{mg\mathrm{tan}\alpha}{\mathrm{cos}\theta +\mathrm{tan}\alpha \mathrm{sin}\theta}\\ \Rightarrow F=\frac{mg\mathrm{tan}\alpha \mathrm{cos}\alpha}{\mathrm{cos}\theta \mathrm{cos}\alpha +\mathrm{sin}\alpha \mathrm{sin}\theta}\\ \Rightarrow F=\frac{mg\mathrm{sin}\alpha}{\mathrm{cos}(\theta -\alpha )}\end{array}$$
The denominator is maximum when cos (θ – α) = 1.
$$\begin{array}{l}\Rightarrow \theta -\alpha =0\\ \Rightarrow \theta =\alpha \end{array}$$
Corresponding minimum force is :
$$\begin{array}{l}\Rightarrow {F}_{\mathrm{min}}=mg\mathrm{sin}\alpha \end{array}$$
Since $\mathrm{tan}\alpha ={\mu}_{s};\phantom{\rule{1em}{0ex}}\mathrm{cos}\alpha =1/\mathrm{sec}\alpha =1/\sqrt{\left(1+\mathrm{tan}{}^{2}\alpha \right)}=1/\sqrt{\left(1+{\mu}_{s}^{2}\right)};$ Hence, $\mathrm{sin}\alpha =\sqrt{\left(1-{\mathrm{cos}}^{2}\alpha \right)}={\mu}_{s}/\sqrt{\left(1+{\mu}_{s}^{2}\right)}$ . Putting this value in the equation,
$$\begin{array}{l}\Rightarrow {F}_{\mathrm{min}}=\frac{{\mu}_{s}mg}{\surd (1+{{\mu}_{s}}^{2})}\end{array}$$
Generally a body is moved either by “pulling” or “pushing” corresponding to the very nature of force, which is defined as a “pull” or “push”. We find that existence of friction against relative motion between two bodies puts certain restriction to the ability of external force to move the body.
This restriction is placed with respect to application of force to move a body over another surface. To illustrate the restriction, we consider a block lying on a rough horizontal surface. An external force, F, is applied on the body to initiate the motion. The figure here shows the free body diagram superimposed on the body diagram.
The x-direction component of the external force parallel to the contact surface tends to move the body. On the other hand, the y-direction component of the external force in the downward direction increases the normal force, which increases the maximum static friction force. Thus, we see that one of the components of applied force tries to move the body, whereas the other component tries to restrict body to initiate motion.
Friction is equal to maximum static friction to initiate motion,
$$\begin{array}{l}{F}_{s}={\mu}_{s}N\end{array}$$
We analyze for zero net force in both component directions to investigate the nature of external force, "F".
$$\begin{array}{l}\sum {F}_{x}=F\mathrm{sin}\theta -{\mu}_{s}N=0\\ \Rightarrow F\mathrm{sin}\theta ={\mu}_{s}N\end{array}$$
and
$$\begin{array}{l}\sum {F}_{y}=F\mathrm{cos}\theta +mg-N=0\\ \Rightarrow N=F\mathrm{cos}\theta +mg\end{array}$$
Combining two equations, we have :
$$\begin{array}{l}\Rightarrow F\mathrm{sin}\theta ={\mu}_{s}(F\mathrm{cos}\theta +mg)\\ \Rightarrow F(\mathrm{sin}\theta -{\mu}_{s}\mathrm{cos}\theta )={\mu}_{s}mg\\ \Rightarrow F=\frac{{\mu}_{s}mg}{\mathrm{sin}\theta -{\mu}_{s}\mathrm{cos}\theta}\end{array}$$
The magnitude of external force is a positive quantity. It is, therefore, required that denominator of the ratio is a positive quantity for the body to move.
$$\begin{array}{l}\Rightarrow \mathrm{sin}\theta -{\mu}_{s}\mathrm{cos}\theta \ge 0\\ \Rightarrow \mathrm{tan}\theta \ge {\mu}_{s}\\ \Rightarrow \theta \ge {\mathrm{tan}}^{-1}\left({\mu}_{s}\right)\\ \Rightarrow \theta \ge \alpha \phantom{\rule{2pt}{0ex}}\text{(angle of friction)}\end{array}$$
The body can be pushed for this condition (θ ≥ α). On the other hand, a force applied within the angle of friction, however great it is, will not be able to push the body ahead. A cone drawn with half vertex angle equal to the angle of friction constitutes a region in which an applied external force is unable to move the body.
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