Question

Consider a bar, of length 3.5 m, shown in the figure, being acted on by three forces, and constrained to rotate about it's left end. The magnitudes of the first two forces are 12 N and 21 N, and the first force is acting on the end of the bar at an angle of 46°

Part (a) What is the torque, in newton-meters, due to F1 on this bar relative to the left end? Use a coordinate system with positive directed out of the screen.

Part (b) What is the torque, in newton-meters, due to F2 on this bar relative to the left end, if this force is acting at the midpoint of the bar? Use a coordinate system with positive directed out of the screen.

Part (c) What is the magnitude of the force F3, in newtons, if the force is a distance 0.25 m from the left end and the bar is not rotating?

Answer 1

a. The torque due to
`\(F_1\)` on the bar relative to the left end is
`\(8.8 \, \text{N}\cdot\text{m}\).`

b. The torque due to
`\(F_2\)` on the bar relative to the left end, when acting at the midpoint, is
`\(0 \, \text{N}\cdot\text{m}\)`.

c. The magnitude of force
`\(F_3\)` required for the bar not to rotate, given a distance of 0.25 m from the left end, is
`\(0 \, \text{N}\)`.

Step-by-step explanation:

a. To calculate the torque
`(\(\tau\))` exerted by a force
`(\(F\))` at an angle
`(\(\theta\))` to the lever arm
`(\(r\))`, we use the formula
`\(\tau = r \cdot F \cdot \sin(\theta)\)`. In this case,
`\(F_1 = 12 \, \text{N}\), \(r_1 = 3.5 \, \text{m}\) (length of the bar), and \(\theta_1 = 46^\circ\).` Plugging in these values, we get:

`\[\tau_1 = 3.5 \, \text{m} \cdot 12 \, \text{N} \cdot \sin(46^\circ) \approx 8.8 \, \text{N}\cdot\text{m}.\]`

Therefore, the torque due to
`\(F_1\) is \(8.8 \, \text{N}\cdot\text{m}\).`

b. When
`\(F_2\)` is acting at the midpoint of the bar (1.75 m from the left end), the lever arm
`(\(r_2\))` becomes 0, as it acts at the point of rotation. Therefore, the torque
`(\(\tau_2\))` can be calculated using
`\(\tau = r \cdot F \cdot \sin(\theta)\),` and since
`\(r_2 = 0\)`, the torque is
`\(0 \, \text{N}\cdot\text{m}\).` The force
`\(F_2\)` does not contribute to the torque about the left end in this case.

c. For the bar not to rotate, the net torque must be zero. The torque
`(\(\tau_3\)) due to \(F_3\) at a distance \(r_3 = 0.25 \, \text{m}\) is given by \(\tau_3 = r_3 \cdot F_3 \cdot \sin(\theta_3)\). Since the bar is not rotating, \(\tau_3\) must be zero. Therefore, \(F_3\) can be calculated as \(F_3 = 0 \, \text{N}\)`to satisfy the equilibrium condition.

Answer 2

To calculate the torques exerted by different forces on a bar, use τ = r * F * sin(θ). Using this formula allows us to find the torque values for forces F1 and F2, and subsequently determine the necessary magnitude for force F3 to keep the bar from rotating.

Step-by-step explanation:

To solve for the torques in these scenarios, we will use the equation for torque, τ = r * F * sin(θ), where τ is the torque, r is the distance from the pivot point, F is the force applied, and θ is the angle between the force vector and lever arm.

Part (a)

The torque due to F1 is calculated using the distance (3.5 m) from the pivot point to where the force is applied and the angle (46°). The formula simplifies to τ = 3.5 m * 12 N * sin(46°). After calculating the sine of 46° and completing the multiplication, we obtain the torque in newton-meters.

Part (b)

The torque due to F2 is calculated at the midpoint of the bar. As the force is perpendicular to the bar and at the midpoint, r = 3.5 m / 2, resulting in τ = (3.5 m / 2) * 21 N * sin(90°), simplifying to torque in newton-meters as the sin(90°) is 1.

Part (c)

The magnitude of force F3 can be found using the fact that the net torque on the bar must be zero for it to not rotate. Assuming the torques exerted by F1 and F2 have been calculated, the torque due to F3 must be equal and opposite to the sum of these torques. Using the distance (0.25 m) from the pivot point to F3 and setting the net torque to 0, we can solve for the magnitude of F3.