Q16Basic Mechanical Engineering

In a V-belt drive, the belt has a trapezoidal cross-section and rides in a V-shaped groove cut into the pulley rim. Unlike a flat belt that rests on the top surface, the V-belt wedges itself into the sides of the groove.

Let $T_1$ be tension on tight side, $T_2$ be tension on slack side, $\mu$ be coefficient of friction, $\theta$ be the angle of contact (in radians), and $2\alpha$ be the total groove angle of the pulley. For a standard flat belt, the tension ratio derived from balancing forces on a small belt element is: $\frac{T_1}{T_2}=e^{\mu \theta }$

However, for a V-belt, the normal reaction ($R_n$) provided by the sides of the groove is much larger than the radial downward force. Resolving the forces vertically in the groove: $R_{radial}=2R_n\sin (\alpha )$ Therefore, the actual normal force causing friction on the sides is $R_n=\frac{R_{radial}}{2\sin (\alpha )}$. The effective total friction force is $F=2\mu R_n=\mu \frac{R_{radial}}{\sin (\alpha )}$. This implies that the effective coefficient of friction is artificially increased to ${\mu }^{\prime }=\frac{\mu }{\sin (\alpha )}$.

Substituting this effective friction into the standard flat belt equation yields the V-belt tension ratio equation:

Because $\sin (\alpha )$ is a fraction (e.g., for a $40^{\circ }$ groove, $\alpha =20^{\circ }$, $\sin (20^{\circ })=0.342$), the exponent becomes nearly three times larger. This exponential increase means a V-belt can transmit massively more power without slipping compared to a flat belt of the same size.