Question

What is the net torque on the bar shown in (Figure 1) about the axis indicated by the dot? Suppose that F=6.0N

Figure is there ^^^

Answer 1

Answer:

net torque on the system is 2.5 Nm

Explanation:

As we know that net torque on the system is given as

$\tau = r_1F_1 - r_2F_2$

now we have

$r_1 = 75 cm$

$F_1 = 6 N$

$r_2 = 25 cm$

$F_2 = 8 N$

now we have

$\tau = 0.75 (6) - 0.25(8)$

$\tau = 2.5 Nm$

Answer 2

The net torque on the bar is 2.5Nm

The formula for calculating torque is expressed as:

$\tau = Fr$

F is the force

r is the perpendicular distance

From the diagram;

Taking the net torque

$\tau_t = F \times 0.75 - 8.0 \times 0.25\\\tau_t=6\times 0.75 -8.0 \times 0.25\\\tau_t = 4.5 - 2.0\\\tau_t = 2.5Nm$

Hence the net torque on the bar is 2.50Nm

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