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By T. Heddle

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E. 1 Ν is the force which gives a 46 FORCE A N D ACCELERATION 47 2 body of mass 1 kg an acceleration of 1 m s~ . Various other systems of units suit eqn. 2), but their use is dying out a n d is not considered here. Solution (a) Assuming a perfectly smooth surface, there is n o opposition to sliding, a n d assuming a perfectly horizontal supporting surface, no other support is required t o oppose the weight of the body. 50 = 10 m s~ . Then, from eqn. 0X10 = 80 N . (a) (b) The friction makes no difference t o the mass of the body but introduces a force acting against the applied force.

E. in the direction of the acceleration. The magnitude of the acceleration a depends upon the magnitude of the resultant force F and u p o n the property of the body called its mass m. T h e relationship between these quantities is given by Newton's second law in the form F oc ma. 2) if the units of F, m, a n d a are such that F = 1 when m = 1 and a = 1. e. 1 Ν is the force which gives a 46 FORCE A N D ACCELERATION 47 2 body of mass 1 kg an acceleration of 1 m s~ . Various other systems of units suit eqn.

2 . 2 . Resolved components of a single force. When only two forces are to be combined (as in Fig. 8), the polygon of forces is a triangle in which the third side represents the resultant. Alternatively, it is sometimes more convenient to draw a parallelogram of forces with a diagonal dividing it into two equivalent triangles. Then the diagonal represents the resultant in either triangle, and the components may be shown running through a common point with the resultant (as in Fig. 9). A right-angled triangle is often used for the opposite purpose, namely that of resolving a given force (resultant) into two perpendicular component forces which are equivalent to the given force.

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