First Year Physics Chapter 2 Notes

• Stay ahead of your peers by using these thorough and well-structured first year physics chapter 2 mcqs.
• Basic concepts of vectors
• What are vectors and Scalars
• Rectangular coordinate system
• Addition of vectors
• resultant vectors
• vector subtraction
• Don't miss out on these free and reliable first year physics chapter 2 short questions study materials for your exams.
• multiplication of a vector by a scalar
• unit vector
• null vector
• equal vectors
• rectangular components of a vector
• Ace your exams with these easy-to-follow first year physics chapter 2 important questions.
• Determination of a vector from its rectangular components
• position vector
• Vector addition by rectangular components: Let A and B be two vectors which are represented by two directed lines OM and ON respectively. The vector B is added to A by the head to tail rule of vector addition (Fig 2.9). Thus the resultant vector R = A + B is given, in direction and magnitude. by the vector OP.
• By Reading first year physics chapter 2 test explore detailed explanations and examples to enhance your understanding.
• In the Fig 2.9 A., B, and R, are the x components of the vectors A, B and R and their magnitudes are given by the lines OQ, MS, and OR respectively, But OROQ + QR OROQMS R, A, B, or OF which means that the sum of the magnitudes of x-components of two vectors which are to be added, is equal to the x-component of the resultant. Similarly the sum of the magnitudes of y-components of two vectors is equal to the magnitude of y-component of the resultant, that is RA, B.
• These first year physics chapter 2 exercise are perfect for quick revisions before your exams.
• The vector addition by rectangular components consists of the following steps.
• (i) Find x and y components of all given vectors.
• (ii) Find x-component R, of the resultant vector by adding the x-components of all the vectors.
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• (ⅲ) Find y-component R, of the resultant vector by adding the y-components of all the vectors.
• (iv) Find the magnitude of resultant vector Rusing R= RR 2
• (v) Find the direction of resultant vector R by using 0=tan R R
• where is the angle, which the resultant vector makes with positive x-axis. The signs of R, and R, determine the quadrant in which resultant vector lies. For that purpose proceed as given below.
• Irrespective of the sign of R, and R,, determine the value of tan from the calculator or by consulting R trigonometric tables. Knowing the value of, angle is determined as follows.
• Product of two vectors
• Scalar product or Dot product
• Vector product or cross-product
• Definition of torque with examples: We have already studied in school physics that a turning. effect is produced when a nut is tightened with a spanner (Fig. 2.13). The turning effect increases when you push harder on the spanner. It also depends on the length of the spanner: the longer the handle of the spanner, the greater is the turning effect of an applied force. The turning effect of a force is called its moment or torque and its magnitude is defined as the product of force F and the perpendicular distance from its line of action to the pivot which is the point O around which the body (spagner) rotates. This distance OP is called moment arm /. Thus the magnitude of torque represented by t is t=IF.
• When the line of action of the applied force passes through the pivot point, the value of moment arm/ = 0, so in this case torque is zero.
• We now consider the torque due to a force F acting on a rigid body. Let the force Facts on rigid body at point P whose position vector relative to pivot O is r. The force F can be resolved into two rectangular components, F sin e perpendicular to r and F cose along the direction of r (Fig. 2.14 a). The torque due to F cos about pivot O is zero as its line of action passes through point O. Therefore, the magnitude of torque due to F is equal to the torque due to F sine only about O. It is given by (Fsine) r=rFsin (2.27) Alternatively the moment arm / is equal to the magnitude of the component of r perpendicular to the line of action of F as illustrated in Fig. 2.14 (b). Thus (rsin) F= rF sin (2.28)
• where is the angle between r and F From Eq. 2.27 and Eq. 2.28 it can be seen that the torque can be defined by the vector product of position vector r and the force F, so T=rxF
• Equilibrium of forces
• The first condition of equilibrium
• Equilibrium of torques
• The second condition of equilibrium

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