---
title: "Static force analysis of slider crank"
subject: "Theory of Mechanism and Machine I"
chapter: 8
key: "semester_6/theory_of_mechanism_and_machine_i:static-force-analysis-slider-crank"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/static-force-analysis-slider-crank
site: BCT Engineering
---

# Static force analysis of slider crank

Theory of Mechanism and Machine I · Chapter 8: Force Analysis of Mechanisms

- Asked in: year not given, 8 marks
- Video: [Static Force Analysis of Slider crank Mechanism](https://www.youtube.com/watch?v=at5NmoCJv4A) — Santosh S Gaidhankar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/static-force-analysis-slider-crank
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-8

## Question

Define two-force member and three-force member and state the conditions of equilibrium for each. In a slider crank mechanism the crank is 100 mm long and the connecting rod is 400 mm long. When the crank makes 45 degrees with the line of stroke, a force of 2000 N acts on the slider along the line of stroke, tending to push it towards the crank. Neglecting weights, inertia and friction, find (a) the force in the connecting rod, (b) the normal reaction of the guide on the slider, (c) the torque on the crank required to keep the mechanism in equilibrium, and verify the torque by the principle of virtual work.

## Short answer

A two-force member has forces at only two points, equal, opposite and collinear. A three-force member has three coplanar, concurrent forces summing to zero. For the slider-crank: connecting rod force = 2032.0 N, guide reaction = 359.2 N, crank torque = 166.8 N m.

## Answer

#### Force members
- **Two-force member**: a member acted upon by forces at only two points (no couple). For equilibrium the two forces are equal, opposite and collinear (along the line joining the points). The connecting rod (with pin joints at both ends and no load between) is a two-force member.
- **Three-force member**: a member with forces at three points. For equilibrium the three forces must be coplanar, their lines of action must meet at one point (concurrent) and their vector sum must be zero (a closed force triangle). The slider block (force $P$, guide reaction $F_N$, rod force $F_Q$) is a three-force member.

#### Geometry
$$
\sin\phi = \frac{r\sin\theta}{l} = \frac{\sin45^\circ}{4} = 0.1768 \Rightarrow \phi = 10.18^\circ
$$

#### Free-body of the slider
```text
          F_N (guide reaction, perpendicular)
            ^
            |
   P  <-----[]----- F_Q (rod force)
```

The rod is a two-force member, so $F_Q$ acts along the rod, at angle $\phi$ to the line of stroke. Equilibrium of the slider:

$$
F_Q\cos\phi = P \Rightarrow F_Q = \frac{2000}{\cos 10.18^\circ} = 2032.0\ \text{N}
$$

$$
F_N = F_Q\sin\phi = P\tan\phi = 2000\tan 10.18^\circ = 359.2\ \text{N}
$$

#### Torque on the crank
The rod force $F_Q$ acts on the crank pin along the rod. The perpendicular distance from the crank centre O to the line of the rod is $r\sin(\theta + \phi)$. So

$$
T = F_Q\, r\sin(\theta + \phi) = 2032.0 \times 0.1 \times \sin 55.18^\circ = 166.8\ \text{N m}
$$

(The radial component on the crank pin, $F_Q\cos(\theta+\phi) = 1160.2$ N, only loads the bearing.)

#### Check by virtual work
Slider position from O: $x = r\cos\theta + \sqrt{l^2 - r^2\sin^2\theta}$. For a virtual displacement $\delta\theta$ of the crank, the work of the input torque equals the work done against $P$:

$$
T\,\delta\theta = P\,\delta x \Rightarrow T = P\left|\frac{dx}{d\theta}\right| = P r\left[\sin\theta + \frac{\sin2\theta}{2\sqrt{n^2 - \sin^2\theta}}\right]
$$

$$
T = 2000 \times 0.1\left[0.7071 + \frac{1.0000}{2 \times 3.937}\right] = 166.8\ \text{N m}
$$

It agrees with the free-body result.

**Answer: $F_Q = 2032.0$ N, $F_N = 359.2$ N, $T = 166.8$ N m.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the equilibrium condition of a two-force member?
   A. The two forces are equal, opposite and collinear
   B. The two forces are perpendicular
   C. The forces are concurrent with a third force
   D. A couple must act
   Answer: A. A two-force member is in equilibrium only if the two forces are equal, opposite and along the line joining their points; the connecting rod is an example. (Key point: Two-force member)

2. For a three-force member in equilibrium:
   A. Forces must be parallel
   B. Forces must be concurrent, coplanar and form a closed force triangle
   C. Two forces must cancel exactly
   D. The couple must be zero only
   Answer: B. Three forces must be coplanar, meet at a point and sum to zero. (Key point: Three-force member)

3. With phi = 10.18 degrees and P = 2000 N, the connecting rod force is:
   A. 2000 sin(phi)
   B. 2000 cos(phi) = 1969 N
   C. P/cos(phi) = 2032 N
   D. P tan(phi) = 359 N
   Answer: C. Equilibrium of the slider along the stroke gives F_Q cos(phi) = P, so F_Q = 2032 N; the guide reaction is P tan(phi) = 359.2 N. (Key point: Rod force)

4. What is the crank torque and how is it found?
   A. F_Q cos(theta+phi) x r = 116 N m
   B. P x r = 200 N m
   C. F_N x r = 35.9 N m
   D. F_Q r sin(theta+phi) = 166.8 N m, which equals P dx/d(theta) by virtual work
   Answer: D. The moment arm of the rod force about O is r sin(theta+phi), and virtual work T = P|dx/dtheta| gives the same 166.8 N m. (Key point: Crank torque)

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