Durham College · MECH 2103

Durham MECH 2103 tutoring

Help with force diagrams, moments and static equilibrium in Durham MECH 2103.

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Course overview

MECH 2103 applies statics to rigid structures in Durham’s Mechanical Engineering Technology curriculum. Forces, moments and free-body diagrams are central, with algebra and trigonometry used to turn the diagram into solvable equations.

MECH 2103 topics to review

Statics of rigid structuresForces, moments and free-body diagramsAlgebraic and trigonometric equilibrium models

Common difficulties

Using a full angled force when only one component creates the moment
Leaving support reactions out of the force balance

Preparing for this course

Review resolving a force into horizontal and vertical components. Mark the force’s point of application and its line of action. For a moment, the relevant distance is perpendicular to the force, or equivalently use the perpendicular force component with the given radius.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with Durham College.

Check the official course catalogue

Choose the component that creates a moment

A weightless horizontal 2 m beam is pinned at its left end. A 300 N downward load acts 1 m from the pin. A cable attached at the right end pulls upward and left at $30^\circ$ to the beam. Find its tension and the pin’s force components.

Show the reasoning

Take moments about the pin
Let $T$ be cable tension and $R_x,R_y$ the horizontal and vertical pin-reaction components, positive rightward and upward. The cable’s vertical component $T\sin30^\circ$ has a 2 m lever arm. Its horizontal component acts along the beam and has zero moment about the pin. Thus $2T\sin30^\circ-300(1)=0$ .
Solve the cable tension
Since $\sin30^\circ=1/2$ , the tension is $T=300\text{ N}$ . Its vertical component is 150 N and its leftward component is $150\sqrt3\text{ N}$ .
Recover the pin reactions
Horizontal balance requires $R_x=150\sqrt3\text{ N}$ rightward. Vertical balance gives $R_y+150-300=0$ , so $R_y=150\text{ N}$ upward. Both force components and the moment now balance.

Try a changed problem

If the cable angle becomes $60^\circ$ , find the new tension and vertical pin reaction.

Check your reasoning

$T=300/(2\sin60^\circ)=100\sqrt3\approx173.2\text{ N}$ . Its vertical component stays 150 N, so $R_y=150\text{ N}$ . The horizontal reaction changes.

How tutoring can help

Start with the problem statement, even if you do not yet have a diagram. A tutor can help you draw the supports and forces, choose a pivot and build the equilibrium equations. We can solve for the reactions together and check the result with a second moment equation.

Exam preparation for MECH 2103

Practise the sequence diagram, axes, equations, solution and independent check. Use moments about more than one point to verify a solved structure.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Independent practice

Free practice and study guides for MECH 2103

40 min

Torque and Static Equilibrium: Find the Perpendicular Lever Arm

Find perpendicular lever arms, choose torque signs and pivots, and check force and torque balance. Three worked examples and six questions with hints and solutions.

14 questions

Right Triangles: Lengths and Angles

Choose Pythagoras or a trigonometric ratio and turn a word problem into a diagram. Eight worked examples and fourteen practice questions, including heights, a ramp, a kite and two triangles in space.

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MECH 2103 · official course reference