Question

A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist your wrist as you grip it between your thumb and forefinger with a force of 5.0 N . Suppose the coefficient of static friction between the knob and your fingers is 0.65.
What is the most torque that you can exert on the knob without having it slip between your fingers?

Answer 1

Answer:

0.04225 Nm

Explanation:

N = Force applied = 5 N

$\mu$ = Coefficient of static friction = 0.65

d = Diameter of knob = 1.3 cm

r = Radius of knob = $\frac{d}{2}=\frac{1.3}{2}=0.65\ cm$

Force is given by

$F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N$

When we multiply force and radius we get torque

Torque on thumb

$\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm$

Torque on forefinger

$\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm$

The total torque is given by

$\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm$

The most torque that exerted on the knob is 0.04225 Nm

Answer 2

The torque is dependent on the force applied to the system. The torque exerted on the knob is 0.04225 Nm.

What is torque?

The torque is defined as the measure of the force that can cause an object to rotate about its axis.

Given that the applied force F is 5 N and the coefficient of static friction $\mu$ between the knob and your fingers is 0.65. The diameter d of the knob is 1.3 cm.

The radius r of the knob is given as,

$r = \dfrac {d}{2}$

$r = \dfrac {1.3}{2}$

$r = 0.65 \;\rm cm$

The force F' applied to the system is calculated as given below.

$F' = F\mu$

$F' = 5\times 0.65$

$F'=3.25 \;\rm N$

Now the torque on the thumb is given as,

$\tau_t = F'\times r$

$\tau_t = 3.25\times 0.65\times 10^{-2}$

$\tau_t = 0.021125 \;\rm Nm$

The torque on the forefinger is,

$\tau_f = F'\times r$

$\tau_f= 3.25\times 0.65\times 10^{-2}$

$\tau_f = 0.021125 \;\rm Nm$

The total torque of the system is calculated as,

$\tau = \tau_t + \tau_f$

$\tau= 0.021125 + 0.021125$

$\tau = 0.04225 \;\rm Nm$

Hence we can conclude that the torque exerted on the knob is 0.04225 Nm.

To know more about the torque, follow the link given below.

https://brainly.com/question/6855614.