By R.K. Shukla
Appropriate for varied aggressive examinations and for the 1st yr engineering scholars, this booklet includes diagrams, worked-out examples and comparable questions-answers.
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Additional info for Mechanics
Example text
To find this instantaneous speed v at a particular time t, we draw a straight-line tangent to the distance- time curve at that value of t. The length of the line does not matter. Then we determine v from the tangent line from the formula v = ∆s/∆t Where ∆s is the distance interval between the ends of the tangent and ∆t is the time interval between them. (∆ is the Greek capital letter delta). The instantaneous speed of the car at t = 40 s is from figure Total distance, (m) 0 100 200 300 400 500 Elapsed time, (S) 0 28 40 49 57 63 V = ∆s/∆t = 100 m/10 s = 10 m/s D ista nce (m ) 5 00 4 00 3 00 2 00 1 00 0 10 20 30 40 50 60 70 Tim e (S ) Fig.
If θ is the angle between F and the direction OX, then OP/OR = cos θ or OP = OR cos θ = F cos θ So the component of any vector F in a direction making an angle θ to F is always given by = F cos θ In a direction OY perpendicular to OX, F has a component F cos (90° – θ) which is F sin θ. This component is represented by OQ in the figure (b). S T O 6N 6 0° 6 0° Y R Q P F F sin θ Q 9 0° θ O F co s θ P X W (a ) (b ) Fig. 1 Components of vectors. 6 PROJECTILE MOTION A body projected in the horizontal direction with a uniform velocity is under the action of (i) uniform velocity in the horizontal direction and (ii) uniform acceleration due to gravity in the vertically downward direction.
This cone is termed “cone of friction” as shown in figure 11. Fig. 11 PROBLEM Q. 1. Two blocks of masses m1 and m2 are connected by a massless spring on a horizontal frictionless table. Find the ratio of their accelerations a1 and a2 after they are pulled apart and then released. 40 Mechanics Solution. The blocks are pulled apart by equal and opposite forces. On being released they start moving toward each other under equal and opposite (elastic) force exerted by the massless spring. Since force is equals to mass multiplied by acceleration, we have F = m1 a1 = m2a2 ∴ a1 m2 = a2 m1 Q.