Compare the magnitudes of the equilibrant vectors measured from the experiment with thoseobtained from the graphical and component methods.Example:A: 200 g 60° above +x axisB: 300 g 45° above -x axisC: 400 g 30° below-x-axisA. A cos a =1.96 N x cos 60°= 0.98 NB = B cos b = 2.94 N x cos 135º = -2.08 NCx = C cos g = 3.92 N x cos 210° = -3.39 NR. A + B + C₂ = -4.49 NA, = A sin a = 1.96 N x sin 60° = 1.70 NBy = B sin b = 2.94 N x sin 135° = 2.08 NCy= C sin g = 3.92 N x sin 210° = -1.96 NRy= Ay+ By + Cy = 1.82 NQuestions:II: (a) A: 150 g 60° along +y axisB: 200 g 45° above -x axisC: 100 g 30° below -x axisA, A cos a =B₂= B cos b =C₂ = C cos g =R-A₂+B+C₂A = A sin a =B, =B sin b =C₂ = C sin g =R₂ = A +By+C₂=R = (R₂. R₂):Quadrant(c)A: 200 g along -y axisB: 100 g 60° above +x axisC: 200 g 45° below →x axisA, A cos a=B₂B cos b =C₂ = C cos g =R₂-A₂+B+C₂A₁ = A sin a =B = B sin b =C₁=C sin g =Ry= Ay+By+Cy=R = (R₁, R₂):Quadrant Questions:II: (a) A: 150 g 60° along +y axisB: 200 g 45° above-x axisC: 100 g 30° below-x axisA = A cos a=B₂= B cos b =C=C cos g =R-A₂+B+C₂A = A sin a =B, =B sin b =C₁=C sin g =R₂= Ay+By+C =R=(R₁, R₂):QuadrantR=√(R₂²+R²) =Direction q = tan¹ [R, R.] =A: 200 g along -y axisB: 100 g 60' above +x axisC: 200 g 45 below -x axisA₂ = A cos a =B, =B cos b=C₁=C cos g =R, A, B, C,A, = A sin a =B, =B sin b =C₁=C sin g =R₂=A₂+B+C =R=(R₂. R₂):________ QuadrantR=√(R₂²+R₂) =Direction:q=tan [R, /R.] =

Given: a) A: 150 g along +y axis B: 200 g 450 above -x axis C: 100 g 300 below -x axis b) A: 200 g…

Questions: II: (a) A: 150 g 60° along +y axis B: 200 g 45° above -x axis C: 100 g 30° below-x axis A = A cos a = B₂= B cos b = C=C cos g = R-A₂+B+C A₂ = A sin a = B, =B sin b = C₁=C sin g = R₂ = Ay+By+C = R=(R₂ R₂): Quadrant R=√(R₂²+R₂²) = Direction:q=tan¹ [R, R.] = A: 200 g alongy axis B: 100 g 60 above +x axis C: 200 g 45 below -x axis A₂ = A cos a = B, =B cos b= C₁=C cos g = R, A, B, C, A₂ = A sin a = B, =B sin b = C₁=C sin g= R₂=A₂+By+C₂ = R=(R₂. R₂):________ Quadrant R= √ (R₂²+R₂²) = Direction:q=tan [R, /R.] =

Questions: II: (a) A: 150 g 60° along +y axis B: 200 g 45° above -x axis C: 100 g 30° below-x axis A = A cos a = B₂= B cos b = C=C cos g = R-A₂+B+C A₂ = A sin a = B, =B sin b = C₁=C sin g = R₂ = Ay+By+C = R=(R₂ R₂): Quadrant R=√(R₂²+R₂²) = Direction:q=tan¹ [R, R.] = A: 200 g alongy axis B: 100 g 60 above +x axis C: 200 g 45 below -x axis A₂ = A cos a = B, =B cos b= C₁=C cos g = R, A, B, C, A₂ = A sin a = B, =B sin b = C₁=C sin g= R₂=A₂+By+C₂ = R=(R₂. R₂):________ Quadrant R= √ (R₂²+R₂²) = Direction:q=tan [R, /R.] =

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter2: Vectors

Section: Chapter Questions

Problem 84AP: A force vector A has x and y components, respectively, of -8.80 units of force and 15.00 units of...

See similar textbooks

Similar questions

For the three-dimensional vectors in the following figure, find (a) GH , (b) GH , and (c) GH .
Four force vectors each have the same magnitude f.What is the largest magnitude the resultant force vector may have when these forces are added? What is the smallest magnitude of the resultant? Make a graph of both situations.
If three vectors sum up to zero, si1at geometric condition do they satisfy?
Vector B is 5.0 cm long and vector A is 4.0 cm long. Find the angle between these two vectors when |A+B|=3.0cm and |AB|=3.0cm .
If A + B = 0, the corresponding components of the two vectors Aand Bmust be (a) equal, (b) positive, (c) negative, (d) of opposite sign.
A force vector A has x and y components, respectively, of -8.80 units of force and 15.00 units of force. The x and y components of force vector B are, respectively, 13.20 units of force and -6.60 units of force. Find the components of force vector C that satisfies the vector equation AB+3C=0 .
Find the angles that vector makes with the and axes.
The following figure shows a triangle formed by the three vectors A , B , and C . If vector C is drawn between the midpoints of vectors A and B , show that C=C/2
Assuming the +x -axis is horizontal and points to the tight, resolve the vectors given In the following figure to their scalar components and express them in vector component form.
Figure 1.19 shows two vectors lying in the xy-plane. Determine the signs of the x- and y-components of A, B, and A+B.
Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]
Using a convenient scale, find the resultant of the following vectors graphically: 10m at40o with the x-axis, 15 m at 140o with the x-axis, and 15 m at 210o with the x-ax. Checkby the component method.