Virtual Work

Classical Mechanics Level 3

A force is being applied by a pneumatic cylinder to a lever, according tot the diagram below.

If $a = 500\text{ mm}$ and $b = 500\text{ mm}$ and, at a specific moment, and $\phi = \frac \pi3$ radians, then calculate the magnitude (neglect + or - sign) of the resulting torque around point C in $\text{Nm}$ .

Details and Assumptions:

Assume points A and C to be fixed points, distances $a$ and $b$ represent solid rods.
Assume the force in the cylinder to be $400\text{ N}$ .
Assume that the construction is frictionless and weightless.

The answer is 100.

1 solution

Kris Hauchecorne
Dec 27, 2016

a = b = 0.5 m

Cosine rule: x² = a² + b² + 2ab sin(phi) (The plus sine is due tot the fact that phi is the outer angle.)

Differentiating for variables x en phi: 2x dx = -2ab sin(phi) d(phi)

Virtual work by force and torque must be equal: F dx = M d(phi), and therfore: F dx / d(phi) = M = -ab sin(phi) F / x

Again from the cosine rule: x² = a² + a² + 2 a² /2 = 3a², so x = sqrt(3) * a

Calculation: M = - (0.5 m * 0.5 m * sqrt(3)/2 * 400 N) / (sqrt(3) * 0.5 m) = -100 Nm

The magnitude of the torque is thus 100 Nm, counter clock wise (if the cilinder is pushing outward).

It's also possible tot simply define vectors r=(1/2; Sqrt(3)/2) 0.5 m and F=(Sqrt(3)/2; 1/2) 400 N and just calculate the outer product, but I like this virtual work twist.

Kris Hauchecorne - 4 years, 5 months ago

Virtual Work

The answer is 100.

1 solution

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