(a) Write each vector in Fig. E1.39 in terms of the unit vectors i^ and j^. (b) Use unit vectors to express vector C, where C=3.00A4.00B (c) Find the magnitude and direction C.

Short Answer

Expert verified

Answer

  1. The representation of A and Bin terms of unit vectors is A=1.23 mi^+3.38 mj^and B=2.08 mi^+1.20 mj^.
  2. The representation of Cin terms of unit vectors is C=12.01 mi^+14.94 mj^.
  3. The magnitude of Cis 19.17 m and its directions is 51.2o in the first quadrant.

Step by step solution

01

 Step 1: Identification of given data

The vector C=3A4B.

02

Step-2: Vector Quantities and their magnitudes.

Consider a Vector quantityR=Rxi^+Ryj^, HereRx,Ryare the components along x,y, directions respectively and i^,j^are the unit vectors along x,y directions respectively. The magnitude of this vector is expressed as,

R=Rx2+Ry2

03

Determination of vectors from the given figure

Part (a)

A can be represented as A=Axi^+Ayj^. Here AX and Ay are the components of A in x and y directions respectively. iand jare the unit vectors in x and y directions respectively.

Using trigonometry, the components of A can be expressed as,

Ax=3.60 mcos70°=1.23 mAy=3.60 msin70°=3.38 m

Thus, A can be represented as,

A=Axi^+Ayj^A=1.23 mi^+3.38 mj^

Similarly, the components of B can be expressed as,

Bx=2.40 mcos30=2.08 mBy=2.40 msin30=1.2 m

Thus, B can be represented as,

B=Bxi^+Byj^B=2.08 mi^+1.20 mj^

Thus, the representation of A and B in terms of unit vectors is A=1.23 mi^+3.38 mj^andB=2.08 mi^+1.20 mj^.

04

Step-4: Representation of Vector C→ in terms of unit vectors

Part (b)

The vector C can be expressed as,

C=3A4B

Substitute 1.23 mi^+3.38 mj^forAand2.08 mi^+1.20 mj^forB

C=3A4B=31.23 mi^+3.38 mj^42.08 mi^+1.20 mj^=3.69 mi^+10.14 mj^8.32 mi^+4.8 mj^=12.01 mi^+14.94 mj^

The representation of Cin terms of unit vectors is C=12.01 mi^+14.94 mj^.

05

Estimation of Magnitude of vector  C→

Part (c)

The Vector C can be expressed as,

C=Cxi^+Cyj^

Substitute 12.01 m for Cx and 14.94m for Cy,

C=12.01 mi^+14.94 mj^

The magnitude of vector C can be expressed as,

C=Cx2+Cy2

Substitute 12.01 m for Cx , and 14.94 m for Cy ,

C=12.01 m2+14.94 m2=367.4 m=19.17 m

The direction can be calculated as,

tanθ=CyCx

Substitute 12.01 m for Cx and 14.94m for Cy .

θ=tan114.94 m12.01 m=51.2°

Thus, the magnitude of C is 19.17 m, and its directions is 51.2o in the first quadrant.

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