Composition of forces
1.Explain what composition (addition) of forces and decomposition (resolution) of a force are!
Solution:
Composition (addition) of forces is replacing a system of forces with a single force — the resultant.
The motion effects on the body do not change.
Decomposition (resolution) of a force is replacing a force with a system of forces that has the same effect.
Two forces acting at one point which
- a) act in the same direction : F = F1 + F2
- b) act in opposite directions : F = F1 – F2
- c) are perpendicular to each other : F2 = F12 + F22
- d) form an angle φ : F2 = F12 + F22 – 2·F1·F2·cos(180o – φ)
Two forces
a) parallel and acting in the same direction
F = F1 + F2
r = r1 + r2
F1·r1 = F2·r2
b) parallel and acting oppositely
F = F1 - F2
r = r1 - r2
F1·r1 = F2·r2
2.Two forces F1 = 8 N, F2 = 6 N act at one point. Calculate their resultant if:
- a.) they act in the same direction
- b.) they act in opposite directions
- c.) they are perpendicular to each other
- d.) they form an angle φ = 60o
Solution:
3.A body of mass 50 kg is acted on by two forces F1 = 25 N, F2 = 35 N at an angle φ = 30o. Determine the acceleration of the body!
Solution:
Analysis:
m = 50 kg, F1 = 25 N, F2 = 35 N, φ = 30o, a = ?
The acceleration of the body is a = 1.16 m·s-2.
4.Determine the magnitude and position of the resultant of two parallel forces F1 = 70 N, F2 = 30 N whose lines of action are at a distance of 0.4 m. The forces act in the same direction.
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5.Determine the magnitude and position of the resultant of two forces F1 = 70 N, F2 = 30 N whose lines of action are at a distance of 0.4 m. The forces act in opposite directions.
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6.Two hunters carry a shot deer of mass 90 kg suspended on a horizontal bar. The distances of the points where the bar is supported by the carriers' shoulders from the point of action of the deer's weight are 0.8 m and 1 m. Calculate the magnitudes of the forces acting on the shoulders of both carriers. Propose a change so that both carriers share the carrying equally.
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