Belt Force
In the figure shown, a pulley is driven by a belt. If the coefficient of friction that prevents slipping between the pulley and belt is 0.3, what counteracting force (N) should be applied so the system is in equilibrium? Ignore centrifugal effects.
Expand Hint
Belt Friction:
$$$F_1=F_2e^{\mu \theta}$$$
where
$$F_1$$
is the applied force in the direction of impending motion,
$$F_2$$
is the applied force resisting impending motion,
$$\mu$$
is the static coefficient of friction, and
$$\theta$$
is the total angle of contact between surfaces in radians.
Hint 2
Unit Circle:
Belt Friction:
$$$F_1=F_2e^{\mu \theta}$$$
where
$$F_1$$
is the applied force in the direction of impending motion,
$$F_2$$
is the applied force resisting impending motion,
$$\mu$$
is the static coefficient of friction, and
$$\theta$$
is the total angle of contact between surfaces in radians. Recall a unit circle to determine
$$\theta$$
:
Therefore,
$$$F_1=(400N)e^{(0.3)\pi}=(400N)(2.566)=1,027\:N$$$
1,027 N
Time Analysis
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