Editing Content :: Admin Scraper

Chapter

Content type

Position

Value

LAWS OF MOTION 65 method in solved examples. To handle a typical the net force on the block must be zero i.e., problem in mechanics systematically, one R = 20 N. Using third law the action of the should use the following steps : block (i.e. the force exerted on the floor by (i) Draw a diagram showing schematically the the block) is equal to 20 N and directed various parts of the assembly of bodies, the vertically downwards. links, supports, etc. (b) The system (block + cylinder) accelerates (ii) Choose a convenient part of the assembly downwards with 0.1 m s . The free-body as one system. diagram of the system shows two forces on (iii) Draw a separate diagram which shows this the system : the force of gravity due to the system and all the forces on the system by earth (270 N); and the normal force R′ by the the remaining part of the assembly. Include floor. Note, the free-body diagram of the also the forces on the system by other system does not show the internal forces agencies. Do not include the forces on the between the block and the cylinder. Applying environment by the system. A diagram of the second law to the system, this type is known as ‘a free-body diagram’. (Note this does not imply that the system 270 – R′ = 27 × 0.1N under consideration is without a net force). ie. R′ = 267.3 N (iv) In a free-body diagram, include information about forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion. (v) If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F. The following example illustrates the above procedure : ⊳Example 4.12 See Fig. 4.15. A wooden Fig. 4.15 block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg By the third law, the action of the system on is placed on top of the block, the floor yields the floor is equal to 267.3 N vertically downward. steadily and the block and the cylinder Action-reaction pairs together go down with an acceleration of For (a): (i) the force of gravity (20 N) on the block 0.1 m s . What is the action of the block on the floor (a) before and (b) after the floor by the earth (say, action); the force of yields ? Take g = 10 m s . Identify the gravity on the earth by the block (reaction) equal to 20 N directed action-reaction pairs in the problem. upwards (not shown in the figure). (ii) the force on the floor by the block Answer (action); the force on the block by the (a) The block is at rest on the floor. Its free-body floor (reaction). diagram shows two forces on the block, the For (b): (i) the force of gravity (270 N) on the force of gravitational attraction by the earth system by the earth (say, action); the equal to 2 × 10 = 20 N; and the normal force force of gravity on the earth by the R of the floor on the block. By the First Law, system (reaction), equal to 270 N, Reprint 2026-27