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LAWS OF MOTION 61 approximately true. Yet they are very useful in Answer The forces acting on a block of mass m practical calculations in mechanics. at rest on an inclined plane are (i) the weight Thus, when two bodies are in contact, each mg acting vertically downwards (ii) the normal experiences a contact force by the other. Friction, force N of the plane on the block, and (iii) the bydefinition,isthecomponentofthecontactforce static frictional forcs f opposing the impending parallel to the surfaces in contact, which opposes motion. In equilibrium, the resultant of these impending or actual relative motion between the forces must be zero. Resolving the weight mg two surfaces. Note that it is not motion, but along the two directions shown, we have relative motion that the frictional force opposes. m g sin θ = fs , m g cos θ = N Consideraboxlyinginthecompartmentofatrain As θ increases, the self-adjusting frictional force that is accelerating. If the box is stationary relative to the train, it is in fact accelerating alongncreases until at θ = θmax, fsachieves its withthetrain.Whatforcescausetheacceleration maximum value, ( s max= µs N. of the box? Clearly, the only conceivable force in the horizontal direction is the force of friction. ITherefore, there were no friction, the floor of the train would –1 slip by and the box would remain at its initial tan θmax = µs or θmax= tan µs position due to inertia (and hit the back side of When θ becomes just a little more than θmax , the train). This impending relative motion is there is a small net force on the block and it opposed by the static friction fs. Static friction begins to slide. Note that θ depends only on provides the same acceleration to the box as that µ and is independent of the mass of the block. s of the train, keeping it stationary relative to the train. For θmax = 15°, ⊳ µs = tan 15° Example 4.7 Determine the maximum = 0.27 ⊳ acceleration of the train in which a box ⊳ lying on its floor will remain stationary, Example 4.9 What is the acceleration of the block and trolley system shown in a given that the co-efficient of static friction between the box and the train’s floor is Fig.4.12(a),ifthecoefficientofkineticfriction 0.15. between the trolley and the surface is 0.04? What is the tension in the string? (Take g = Answer Since the acceleration of the box is due -2 10 m s ). Neglect the mass of the string. to the static friction, ma = f ≤ µ N = µ m g s s s i.e. a ≤ µs g ∴ amax= µsg = 0.15 x 10 m s –2 = 1.5 m s –2 ⊳ ⊳ Example 4.8 See Fig. 4.11. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an anglθ = 15° with the horizontal, the mass just begins to slide.Whatisthecoefficientofstaticfriction between the block and the surface ? (a) (b) (c) Fig. 4.11 Fig. 4.12 Reprint 2026-27