Revisions to Is friction always opposite to velocity?

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The contact force referred to as kinetic friction is applied by one surface on another in the direction opposite to that of the relative velocity of the latter surface w.r.t. the former surface.
The contact force referred to as static friction is is applied by one surface on another in the direction opposite to that of the 'impending' relative velocity of the latter surface. The direction of the impending velocity, which is a fictitious quantity, is in the direction of the relative acceleration (w.r.t. the surface applying the force) of the point of contact resulting from the dynamics in which the friction force of interest is fictitiously assumed to be vanished w.r.t. the former surface.

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edited May 1, 2021 at 18:01

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Force driven wheel (figure on the left): The Newton's laws of motion imply that $F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$ and $-fR=I\alpha$ which implies that $0\leq a$, $\alpha\leq 0$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $F$ which points in the $+X$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $F=f$. The static friction condition $f\leq \mu_s W$ therefore implies that $F=3f\leq \mu_s W=3\mu_s mg$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{2f}{m}$ allowable under the rolling without slipping regime. In fact, this is the underlying reason why circular wheels are more efficient than non-circular ones.
Torque driven wheel (figure on the right): The Newton's laws of motion imply that $-\tau+fR=I\alpha$ and $f=\frac{W}{g}a=-\frac{W}{g}\alpha R$, which implies that $0\leq a$, $\alpha\leq 0$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $F$, that is in the direction $-X$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $\tau=-\frac{3}{2}{f}{R}$. The static friction condition $f\leq \mu_s W$ therefore implies that $\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{f}{m}$ allowable under the rolling without slipping regime.

Force driven wheel (figure on the left): The Newton's laws of motion imply that $F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$ and $-fR=I\alpha$ which implies that $0\leq a$, $\alpha\leq 0$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $F$ which points in the $+X$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $F=f$. The static friction condition $f\leq \mu_s W$ therefore implies that $F=3f\leq \mu_s W=3\mu_s mg$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{2f}{m}$ allowable under the rolling without slipping regime.
Torque driven wheel (figure on the right): The Newton's laws of motion imply that $-\tau+fR=I\alpha$ and $f=\frac{W}{g}a=-\frac{W}{g}\alpha R$, which implies that $0\leq a$, $\alpha\leq 0$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $F$, that is in the direction $-X$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $\tau=-\frac{3}{2}{f}{R}$. The static friction condition $f\leq \mu_s W$ therefore implies that $\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{f}{m}$ allowable under the rolling without slipping regime.

Force driven wheel (figure on the left): The Newton's laws of motion imply that $F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$ and $-fR=I\alpha$ which implies that $0\leq a$, $\alpha\leq 0$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $F$ which points in the $+X$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $F=f$. The static friction condition $f\leq \mu_s W$ therefore implies that $F=3f\leq \mu_s W=3\mu_s mg$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{2f}{m}$ allowable under the rolling without slipping regime. In fact, this is the underlying reason why circular wheels are more efficient than non-circular ones.
Torque driven wheel (figure on the right): The Newton's laws of motion imply that $-\tau+fR=I\alpha$ and $f=\frac{W}{g}a=-\frac{W}{g}\alpha R$, which implies that $0\leq a$, $\alpha\leq 0$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $F$, that is in the direction $-X$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $\tau=-\frac{3}{2}{f}{R}$. The static friction condition $f\leq \mu_s W$ therefore implies that $\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{f}{m}$ allowable under the rolling without slipping regime.

edited text for better explanation

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edited May 1, 2021 at 10:50

kbakshi314

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Force driven wheel (figure on the left): The Newton's laws of motion imply that $F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$ and $-fR=I\alpha$ which implies that $0\leq a$, $\alpha\leq 0$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $F$ which points in the $+X$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $F=f$. The static friction condition $f\leq \mu_s W$ therefore implies that $F=3f\leq \mu_s W=3\mu_s mg$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{2f}{m}$ allowable under the rolling without slipping regime.
Torque driven wheel (figure on the right): The Newton's laws of motion imply that $-\tau+fR=I\alpha$ and $f=\frac{W}{g}a=-\frac{W}{g}\alpha R$, which implies that $0\leq a$, $\alpha\leq 0$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $F$, that is in the direction $-X$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $\tau=-\frac{3}{2}{f}{R}$. The static friction condition $f\leq \mu_s W$ therefore implies that $\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{f}{m}$ allowable under the rolling without slipping regime.

Force driven wheel (figure on the left): The Newton's laws of motion imply that $F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$ and $-fR=I\alpha$ which implies that $0\leq a$, $\alpha\leq 0$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $F$ which points in the $+X$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $F=f$. The static friction condition $f\leq \mu_s W$ therefore implies that $F=3f\leq \mu_s W=3\mu_s mg$, which provides the upper bound on the driving force which allows rolling without slipping.
Torque driven wheel (figure on the right): The Newton's laws of motion imply that $-\tau+fR=I\alpha$ and $f=\frac{W}{g}a=-\frac{W}{g}\alpha R$, which implies that $0\leq a$, $\alpha\leq 0$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $F$, that is in the direction $-X$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $\tau=-\frac{3}{2}{f}{R}$. The static friction condition $f\leq \mu_s W$ therefore implies that $\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$, which provides the upper bound on the driving torque which allows rolling without slipping.

Force driven wheel (figure on the left): The Newton's laws of motion imply that $F-f=\frac{W}{g}a=-\frac{W}{g}\alpha R$ and $-fR=I\alpha$ which implies that $0\leq a$, $\alpha\leq 0$. We observe that if the direction of the static friction force is reversed, we would obtain a contradiction since the rolling without slipping condition would be violated (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in the direction of the applied force $F$ which points in the $+X$ direction, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $F=f$. The static friction condition $f\leq \mu_s W$ therefore implies that $F=3f\leq \mu_s W=3\mu_s mg$, which provides the upper bound on the driving force which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{2f}{m}$ allowable under the rolling without slipping regime.
Torque driven wheel (figure on the right): The Newton's laws of motion imply that $-\tau+fR=I\alpha$ and $f=\frac{W}{g}a=-\frac{W}{g}\alpha R$, which implies that $0\leq a$, $\alpha\leq 0$. Clearly, assuming that the direction of the friction of force is opposite to that shown in the figure will lead to a contradiction violating the rolling without slipping condition (because the direction of acceleration obtained would be opposite to that required in the known relationship $\vec{a}=-\vec{\alpha}\times R\hat{k}=-\alpha \hat{j} \times R\hat{k}$ to obtain the correct rightward acceleration of the COM). Notice that the direction of the 'impending' relative velocity of the point at the location of surface contact on the body (w.r.t. the ground surface) is in opposite to the direction of the applied force $F$, that is in the direction $-X$, and that the force of friction acts opposite to this direction. Further, as an aside, notice that if the circular body is a uniformly dense cylinder of mass $m:=\frac{W}{g}$, then $I=m\frac{R^2}{2}$, so that the equations of motion yield $\tau=-\frac{3}{2}{f}{R}$. The static friction condition $f\leq \mu_s W$ therefore implies that $\tau\leq \frac{3}{2}\mu_s WR=\frac{3}{2}\mu_s mgR$, which provides the upper bound on the driving torque which allows rolling without slipping. Finally, the derived bound provides insight into the upper bound of acceleration $a=\frac{f}{m}$ allowable under the rolling without slipping regime.

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edited May 1, 2021 at 10:44

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edited Apr 28, 2021 at 17:59

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edited Apr 27, 2021 at 22:58

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