Statics · Concurrent forces
Condition: The theoretical foundations explain free-body diagrams, force polygons and equilibrium using a wall bracket as an example. Exercise 1 builds on this example.
Open theoretical foundations →First solve each task the classic way on paper: sketch the system, draw the free-body diagram, set up the balances and solve by hand — exactly as in the exam later. The subtasks on this page are your check: enter your results and use them to check every step of your hand calculation individually.
Exercise 1 · Concept check
How do you recognise in the force polygon that a concurrent force system is in equilibrium?
Equilibrium means — chained tip to tail, the last arrow lands back at the start of the first: the force polygon is closed.
Match the terms to their descriptions.
All lines of action pass through one point
Bar force positive under the tension assumption — the bar is stretched
Bar force negative under the tension assumption — the bar braces
Graphical vector addition, tip to tail
Concurrent = one common intersection point of the lines of action (no moments) · tension bar positive, compression bar negative (under the tension assumption) · the force polygon adds vectors tip to tail.
Compare this bracket with the wall bracket in chapters 01 and 02 of the theoretical foundations: here, wall connection B of the inclined member is above connection A rather than below it. Practise isolating a joint: choose the point where both member forces and the load form a concurrent force system, and draw the three forces there. You may assume tension or compression for the member forces. Consider how the direction of the inclined member and its tension or compression differ from the theory example.
What stays the same is the procedure: In the joint free-body diagram practised here, joint K is isolated from both members and the load suspension. The load acts vertically downwards; the member forces act along their members and, when tension is assumed, point away from the node. The inclined bar now runs to the left top, so there.
What changes is the Result: because the inclined bar now holds the load from above, it has to pull — its vertical component carries , so (tension member). The horizontal bar is thereby pushed away from the wall: (compression member). Compared with the wall bracket in the theoretical foundations tension and compression swap bars — the same calculation with the signs reversed.
Conclusion: two tools, one principle
Graphically (closed force polygon) and computationally () are the same statement — the next two tasks use both in the three-step solution path.
Task 2 · Traffic light on two ropes
A traffic light with the weight hangs centred between two masts. Both ropes rise at the same angle to the horizontal up to the masts.
Start with the free-body diagram: which three forces act on traffic-light joint K, and what does “a rope can only pull” mean for their directions?
Resolve each rope force into its horizontal and vertical component and balance the two directions separately.
Isolate traffic-light joint K from both ropes and the load suspension, then draw the three forces. Remember: a rope can only pull — unlike the bars, the sense of direction is therefore not free here.
In the joint free-body diagram practised here, traffic-light joint K is isolated from both ropes and the load suspension. The isolated joint is acted on by the load acts vertically downwards and both rope forces act along the ropes away from the node — ropes always pull.
First, only the horizontal direction: which components of the three forces act in the x direction? Choose the correct term.
Horizontally the cos components count: rope 1 pulls to the left (−), rope 2 to the right (+). The load hangs vertically — it does not appear in the x balance at all.
Now the vertical direction — choose the correct term.
Vertically, both sin components carry the load together: . Both ropes pull upwards (+), the load downwards (−).
Look at your x balance from part b: . What does that imply for the two rope forces?
Both terms contain the same factor — cancelling yields . The symmetry of the arrangement is built right into the equation. From now on all it takes is one unknown .
Substitute into your y balance from part c and solve for . Enter the formula symbolically.
Notation: times * · divided / · plus/minus + - · to the power of ^ · square root sqrt() — symbols of this task: G, alpha, functions sin cos tan
Now the numerical value: substitute and into your formula from part e. Set the calculator to Degree and give the result to two decimal places.
Conclusion: the shallow angle drives the force
The rope force (142 N) is clearly larger than half the load (60 N) — at shallow angles increases sharply. That is why a taut rope can never be pulled perfectly straight.
Task 3 · Asymmetric suspension
A lantern with hangs on two ropes: the left one rises at , the right one at to the horizontal. Both rope forces are to be determined.
Two equations, two unknowns: which of the two balances can be solved for one unknown without knowing the other?
Unequal angles mean unequal rope forces — the symmetry from exercise 2 does not hold here. At the end, check which rope carries more and whether that fits the geometry.
Isolate lantern joint K and draw all three forces with their directions. Observe the different rope angles.
The circular cut around K separates both ropes and the lantern suspension. S₁ pulls along the left rope at 30° upwards to the left, S₂ along the right rope at 45° upwards to the right. Both arrows point away from the node: ropes can only pull. G acts vertically downwards.
First, only the horizontal direction — choose the correct term. Careful: here the ropes have different angles.
Rope 1 pulls up and to the left (−cos α), rope 2 up and to the right (+cos β): .
Now the vertical direction — choose the correct term.
Both sin components carry the load together: .
Two equations, two unknowns. First step towards solving: rearrange your x balance from part b for — then only one unknown will remain.
Notation: times * · divided / · plus/minus + - · to the power of ^ — symbols of this task: S1, alpha, beta, functions sin cos tan
Substitute your from part d into the y balance from part c and solve for . Simplify using the addition theorem and enter the formula symbolically.
Notation: times * · divided / · plus/minus + - · to the power of ^ · square root sqrt() — symbols of this task: G, alpha, beta, functions sin cos tan
Numerical value for : insert , and into your formula from part e (calculator set to Degree).
And the numerical value for : substitute your result from part f into the relation from part d.
Conclusion: the steeper rope carries more
The right rope (β = 45°) carries 179 N, more than the left one (146 N). The reason lies in the x balance: both ropes must supply the same horizontal component, so and hence . Because of its smaller cosine the steeper rope needs the larger force — that it thereby also carries the larger share of the load is the consequence, not the cause. The check works out.
Report
The completion status of all three tasks is exported as a PDF report with a verification token — checkable via the platform's report verification.
| Exercise | Status |
|---|---|
| Exercise 1: Concept check | Open |
| Task 2: Traffic light on two ropes | Open |
| Task 3: Asymmetric suspension | Open |