Question

A 46.0-g meter stick is balanced at its midpoint (50.0 cm, zero point is a left end of stick). Then a 210.0-g weight is hung with a light string from the l2 =70.0-cm point, and a 105-g weight is hung from the l1 = 10.0-cm point. Calculate the clockwise and counterclockwise torques acting on the board due to the four forces about the following axes: Calculate the clockwise torque if the axis is the 50-cm point. (Express your answer to three significant figures.) Calculate the counterclockwise torque if the axis is the 50-cm point. (Express your answer to three significant figures.) Calculate the clockwise torque if the axis is the 100-cm point. (Express your answer to three significant figures.) Calculate the counterclockwise torque if the axis is the 100-cm point. (Express your answer to three significant figures.)

Answer 1

Clockwise torque due to 100g is 0.1029 Nm and 200g is 1.4406 Nm. Clockwise torque due to stick mass is 0.2254 Nm and Counter-clockwise torque due to normal force is 1.7689 Nm.            

What is clockwise torque?

The right-hand rule for cross products determines the direction of torque, which is calculated as the cross product of force and distance. Your thumb will point in the direction of the torque if you place your palm in the direction of the applied force and extend your fingers from the pivot point in that direction.

A related right-hand rule relates the direction of the rotation to the direction of the torque. Your fingers will curl in the direction of rotation if you point your thumb in the direction of the torque.

Positive torques cause counter clockwise rotation, while negative torques cause clockwise rotation.

The sum of all torques must be zero at equilibrium since an object in equilibrium has no net torque.

When the force is applied in a direction perpendicular to the line connecting the pivot and the force, the torque is at its greatest.

You can calculate the torque's magnitude using

                                             $\begin{displaymath}\tau =rF_{\bot }=rF\sin \theta .\end{displaymath}$

To solve problems involving torques, follow these eight steps: read the issue, create a free-body diagram, locate the pivot point, write down the expressions for all torques, For equilibrium conditions, set the sum of torques to zero, list all known variables, pick the desired variable(s), write down equations involving those variable(s), solve the equations, plug in numbers, and test your solution.

Clockwise torque due to 100 g                                                                         ⇒ T1 = 0.105* 9.8* 0.1 = 0.1029 Nm

Clockwise torque due to 200 g                                                                                                      ⇒ T2 = 0.210* 9.8* 0.7 = 1.4406 Nm

Clockwise torque due to stick mass                                                                               ⇒ T3 = 0.046* 0.5* 9.8 =0.2254 Nm

Counter-clockwise torque due to normal force                                                                             ⇒ T4 = (0.046 + 0.21 + 0.105)*9.8* 0.5 = 1.7689 Nm

Learn more about torque

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