Midterm III Practice

This collection features solved problems covering work and energy, linear momentum and its conservation, and rotational dynamics—including rotational motion, moments of inertia, and angular momentum.

---

Problem 2 — The Hinge-and-Cable Crane

This exercise focuses on applying rotational equilibrium (torque) and translational equilibrium to a hinged beam system, emphasizing strategic pivot-point selection to simplify solving for unknown forces.

ExercisePromptSolution

A uniform horizontal construction beam of mass \(M = 8\text{ kg}\) and length \(L = 5.0\text{ m}\) is attached to a vertical wall by a frictionless hinge at Point O (the origin, \(x = 0\)). The beam is held horizontal by a cable attached to its far right end (Point E, \(x = 5.0\text{ m}\)). The cable makes an angle of \(\theta = 37^\circ\) relative to the horizontal beam.

A cargo sign of mass \(m = 2\text{ kg}\) hangs from the very end of the beam at Point E.

(Use \(g = 10\text{ m/s}^2\), \(\sin(37^\circ) = 0.6\), and \(\cos(37^\circ) = 0.8\).)

Part A: Torque as a Vector Cross Product

1. Focus on the tension force \(\vec{T}\) acting at Point E. Define the position vector \(\vec{r}\) from the pivot O to the point of application E.
   • Calculate the torque vector \(\vec{\tau}_T\) produced by the cable about Point O using the formal cross-product magnitude formula: \(\tau = rF\sin\phi\).
   • Identify the direction of this torque vector using the Right-Hand Rule (RHR) (i.e., into the page or out of the page).
   • Calculate this same torque using the lever-arm (perpendicular distance) method: \(\tau = r_{\perp}F\). Show that both methods yield the same algebraic expression.

Part B: Solving Equilibrium & Bypassing Forces

At the hinge (Point O), there is an unknown hinge force with horizontal and vertical components (\(H_x\) and \(H_y\)).

1. To find the tension \(T\) in the cable without needing to know anything about the hinge forces, where is the most beneficial place to set your pivot point? Use this pivot point to calculate the tension \(T\) in the cable.

2. Imagine you are asked to find the vertical hinge force \(H_y\) directly, without calculating the cable tension \(T\) first.

   • Explain why choosing Point E (the right end of the beam) as your pivot point allows you to do this.
   • Calculate \(H_y\) using this pivot point (\(\Sigma \tau_E = 0\)).

3. Verify your value for \(H_y\) and find \(H_x\) using standard translational equilibrium equations (\(\Sigma F_x = 0\) and \(\Sigma F_y = 0\)).

I need an exercise for exam practice for a General Physics I lecture at the college level (algebra-based).

- Topic: Equilibrium with no net forces and no net torques.
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Cross products to calculate torques, and beneficial choices of the rotational axis.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Part A: Torque as a Vector Cross Product

1. Analyzing torque on the cable

   • Position vector: \(\vec{r}\) starts at the pivot (O) and points to the right end (E). Its magnitude is \(r = 5.0\text{ m}\).

   • Torque formula (\(\tau = rT\sin\phi\)):

     The angle between the position vector \(\vec{r}\) (pointing right) and the tension force \(\vec{T}\) (pointing up and left) is \(\phi = 180^\circ - 37^\circ\).

     Trig identity check: \(\sin(180^\circ - \theta) = \sin\theta\). Thus, \(\sin(143^\circ) = \sin(37^\circ) = 0.6\).

     \[\tau_T = rT\sin(37^\circ) = (5.0\text{ m}) \cdot T \cdot (0.6) = 3T\]

   • Direction (Right-Hand Rule):

     Point the fingers of your right hand along \(\vec{r}\) (to the right), then curl them in the direction of \(\vec{T}\) (upwards and leftwards). Your thumb points out of the page (positive \(z\)-direction, \(+z\)). This torque tends to rotate the beam counter-clockwise (CCW).

   • Lever-arm method (\(\tau = r_{\perp}T\)):

     Draw a line along the force vector \(\vec{T}\) (line of action). The perpendicular distance from the pivot O to this line is:

     \[r_{\perp} = r\sin(37^\circ) = 5.0 \times 0.6 = 3.0\text{ m}\]
     \[\tau_T = r_{\perp}T = 3T\]

     Both methods yield the exact same result: \(\tau = 3T\).

Part B: Solving Equilibrium & Bypassing Forces

1. Finding tension \(T\)

   • Beneficial pivot: Point O (the hinge).

   • Why: The hinge forces \(H_x\) and \(H_y\) act directly at \(r = 0\). By placing the pivot at Point O, their torques are exactly zero (\(\tau = 0 \cdot F = 0\)), completely eliminating these two unknowns from the torque equation.

   Set up \(\Sigma \tau_O = 0\) (taking CCW as positive):

   • Torque of beam weight (\(W_{\text{beam}} = Mg = 80\text{ N}\) acting at the center of gravity, \(r = 2.5\text{ m}\)):

     \[\tau_{\text{beam}} = -W_{\text{beam}} \cdot \left(\frac{L}{2}\right) = -80 \times 2.5 = -200\text{ N}\cdot\text{m} \quad \text{(CW)}\]

   • Torque of hanging sign (\(W_{\text{sign}} = mg = 20\text{ N}\) acting at the far end, \(r = 5.0\text{ m}\)):

     \[\tau_{\text{sign}} = -W_{\text{sign}} \cdot L = -20 \times 5.0 = -100\text{ N}\cdot\text{m} \quad \text{(CW)}\]

   • Torque of cable:

     \[\tau_T = +3T \quad \text{(CCW)}\]
   \[\Sigma \tau_O = 3T - 200 - 100 = 0\]
   \[3T = 300 \implies T = 100\text{ N}\]

2. Bypassing tension: pivot at Point E

   • The concept: If you place your pivot at Point E (the right end), any force acting at Point E has a position vector of \(\vec{r} = 0\). Therefore, both the cable tension \(T\) and the hanging sign weight \(W_{\text{sign}}\) produce zero torque.

   • Additionally, the horizontal hinge force \(H_x\) points directly along the beam toward Point E, meaning its angle is \(180^\circ\) (\(\sin(180^\circ) = 0\)), so its torque is also zero.

   • This leaves only \(H_y\) and the beam's weight \(W_{\text{beam}}\) to produce torque.

   Set up \(\Sigma \tau_E = 0\) (taking CCW about Point E as positive):

   • The beam weight \(W_{\text{beam}} = 80\text{ N}\) acts at a distance of \(2.5\text{ m}\) to the left of Point E. Pulling down, it tends to rotate the beam CCW around Point E.

     \[\tau_{\text{beam}, E} = +(80\text{ N})(2.5\text{ m}) = +200\text{ N}\cdot\text{m}\]

   • The vertical hinge force \(H_y\) acts at a distance of \(5.0\text{ m}\) to the left of Point E. Pushing upward, it tends to rotate the beam CW around Point E.

     \[\tau_{H_y, E} = -H_y(5.0\text{ m})\]
   \[\Sigma \tau_E = 200 - 5.0H_y = 0\]
   \[5.0H_y = 200 \implies H_y = 40\text{ N}\]

   This bypassed the tension calculation entirely to find \(H_y\) in a single elegant step.

3. Verification via translational forces

   Let's check if our values are consistent using linear force balancing:

   • Vertical forces (\(\Sigma F_y = 0\)):

     \[H_y + T\sin(37^\circ) - W_{\text{beam}} - W_{\text{sign}} = 0\]
     \[40\text{ N} + (100\text{ N})(0.6) - 80\text{ N} - 20\text{ N} = 0\]
     \[40 + 60 - 100 = 0 \quad \text{(Checks out!)}\]

   • Horizontal forces (\(\Sigma F_x = 0\)):

     The only horizontal forces are the horizontal hinge reaction force \(H_x\) (pointing right) and the horizontal component of the cable tension (pointing left):

     \[H_x - T\cos(37^\circ) = 0\]
     \[H_x = T\cos(37^\circ) = (100\text{ N})(0.8) = 80\text{ N}\]

Feedback & Contributions

Have feedback, want to share your experience, or found a fun variation of this exercise? Email jens.f.mahlmann@dartmouth.edu — we'd love to hear from you!