Magnitude of torque can be found by relating the amount of force applied perpendicular to a lever arm about a point of rotation.

$\displaystyle \tau=rF$

In this case, the force is not perpendicular, so we must take the perpendicular aspect of the force to find torque.

$\displaystyle \tau=rFsin\theta$

Plug in and solve.

$\displaystyle \tau=1.5m(100N)sin(30^{\text o})$

$\displaystyle \tau=75Nm$

To calculate the torque experienced by the object, we must take the cross product of the force vector and the position vector.

$\displaystyle \vec{\tau}=\vec{r}\times\vec{F} = \begin{vmatrix} \hat{i}& \hat{j} & \hat{k}\\ r_x&r_y &r_z \\ F_x&F_y &F_z \end{vmatrix}= \hat{i}\left(r_yF_z-r_zF_y \right )-\hat{j}\left(r_xF_z-r_zF_x \right )+\hat{k}\left( r_xF_y-r_yF_x\right )$

$\displaystyle =-10\hat{k} Nm$

The definition of torque is

$\displaystyle T=F\cdot d\cdot sin(\theta)$

 Where

$\displaystyle F$ is the force

$\displaystyle d$ is the distance

Theta is the angle between the direction of the force and the distance.

In this case, $\displaystyle \theta =90^{\circ}$, so $\displaystyle sin(\theta)=1$ .

Plugging in our remaining values:

$\displaystyle T=10N\cdot.15\cdot1$

$\displaystyle 1.5 Newton\cdot Meters=T$

$\displaystyle \Delta L=\tau*\Delta t$

$\displaystyle \Delta L=L_f-L_i$

Initial angular momentum is zero

$\displaystyle \Delta L=L_f$

Combine equations:

$\displaystyle L_f=\tau*\Delta t$

Solve for $\displaystyle \tau$

$\displaystyle \frac{L_f}{\Delta t}=\tau$

Definition of $\displaystyle L$:

$\displaystyle L=I\omega$

$\displaystyle I=.5mr^2$

Combine equations:

$\displaystyle L=.5mr^2\omega$

$\displaystyle \frac{.5mr^2\omega_f}{\Delta t}=\tau$

Plug in values:

$\displaystyle \frac{.5*75*1^2*2\pi}{2.5}=\tau$

$\displaystyle 94.2N\cdot m$

The torque $\displaystyle \tau$ is produced by a force $\displaystyle F$ acting at a radius $\displaystyle R$. Since the force and the radius are perpendicular, then the torque equation gives us:

 $\displaystyle \tau=RF\sin(90^{\circ})=RF$

The new torque, which we will call $\displaystyle \tau_n$ is produced by a force of $\displaystyle 2F$ acting at a radius of $\displaystyle \frac{3R}{4}$ at an angle of . Thus the torque equation gives us:

$\displaystyle \tau_n=(2F)(\frac{3R}{4})\sin(30^\circ)=(2F)(\frac{3R}{4})(\frac{1}{2})=\frac{3}{4}RF$

Since $\displaystyle \tau=RF$, plugging this in to the above gives us $\displaystyle \tau_n=\frac{3}{4}\tau$

Torque is given by:

$\displaystyle \tau=rFsin(\theta)$, where $\displaystyle r$ is the length of the rod from the pivot point, $\displaystyle F$ is the force acting on the rod, and $\displaystyle \theta$ is the angle. Since we have all of these components, we can plug in and solve:

$\displaystyle \tau=2N*5m*sin(90^o)=10J$

A system will only have a net torque of 0 when it has no net force or the net force goes through its center of gravity. The only forces applied in this system is from gravity. Therefore, we need to find the orientation at which all gravitational force goes through point p. This occurs when the rod is oriented vertically, thus $\displaystyle \angle c = 90^{\circ}$. 

Note: $\displaystyle \angle c=270^{\circ}$ will also result in a net torque of 0. At this orientation, the rod is also vertical with the masses in swapped positions.

The equation for torque is $\displaystyle rFsin\theta$. Looking at this equation we can infer that the maximum distance from the center would give maximum torque. Also, any angle besides $\displaystyle 90^{\circ}$ or $\displaystyle 270^{\circ}$ will give an absolute value of a number below one, giving us a smaller torque.

To calculate the torque on the pendulum, we need to know the position of the pendulum. We can find this using the following expression:

$\displaystyle \theta = \theta _{max}cos\left ( \sqrt{\frac{L}{g}}t\right )$

Note that we are using the cosine function because the pendulum begins at it's maximum angle. Plugging in our values:

$\displaystyle \theta =90^{\circ}cos\left ( \sqrt{\frac{(4m)}{\left ( 10\frac{m}{s^2}\right )}}(2s)\right )$

$\displaystyle \theta = 90^{\circ}$

The pendulum is still horizontal, but now on the other side. Now we can directly calculate the torque placed on the pendulum

$\displaystyle \tau = F\cdot r$

Where the radius is the length of the pendulum and the force is the weight of the block (since the pendulum is horizontal).

$\displaystyle \tau = mgL$

$\displaystyle \tau = (2kg)\left ( 10\frac{m}{s^2}\right )(4m)$

$\displaystyle \tau = 80N\cdot m$

For this question, we're given a number of scenarios in which an equal magnitude of force is applied to a rotating bar at a variety of different orientations and locations on the bar. We're then asked to identify which one would generate the most amount of torque.

First, let's recall that torque is a twisting force. That is, it is a force that causes an object to rotate about a pivot point, such as a sea-saw. We can write an expression for torque as follows.

$\displaystyle \tau=Frsin(\Theta )$

Where $\displaystyle F$ is the magnitude of the applied force, $\displaystyle r$ is the distance of the applied force from the pivot point, and $\displaystyle \Theta$ is the angle between the applied force vector and the surface upon which the force is being applied.

Sometimes, the equation for torque is also expressed as follows.

$\displaystyle \tau=Fl$

Where $\displaystyle l$ stands for the lever arm, which takes into account both $\displaystyle r$ and $\displaystyle \Theta$. Thus, for any given magnitude of force, the torque will be the highest when $\displaystyle r$ is greater and when $\displaystyle \Theta$ approaches $\displaystyle 90^{o}$.

With this expression in mind, we can look at each image and make a qualitative assessment of which one will have the greatest torque. We're looking for a diagram in which the force vector is furthest from the pivot point, and is also oriented as close to $\displaystyle 90^{o}$ with respect to the surface of the rotating object. This situation is described by the following picture, thus making it the correct answer.