EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 7, Problem 41PQ

Three billiard balls, the two-ball, the four-ball, and the eight-ball, are arranged on a pool table as shown in Figure P7.26. Given the coordinate system shown and that the mass of each ball is 0.150 kg, determine the gravitational field at x = 2 m , y = 0 .

Expert Solution & Answer
Check Mark
To determine

The gravitational field at x=2cm, y=0.

Answer to Problem 41PQ

The gravitational field at x=2cm, y=0 is (4×1012i^+1×1011j^)m/s2 .

Explanation of Solution

The following figure gives the direction of field vectors.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 7, Problem 41PQ

The point at which gravitational field is to be calculated is identified as (2m,0m).

Use figure1, calculate the distance between two-ball and point at which field has to be calculated.

  r2=(1m)2+(2m)2=5m

Here, r2 is the distance between two-ball and point at which field is to be calculated.

Use figure1, to calculate angle θ.

  tanθ=1m2mθ=tan1(0.5)=27°

Here, θ is the angle that gravitational field due to two-ball at x=2cm, y=0 makes with x axis.

Write the expression for the gravitational field at a point due to a mass m.

  g=Gmr2r^ (I)

Here, g is the gravitational field at a point, G is the universal gravitational constant, m is the mass of the object which produce the field, r^ is the unit vector along the direction of field and r is the distance between the object and the point at which field is to be calculated.

Use equation (I) to write expression for the gravitational field at (2m,0m) by eight-ball.

    g8=Gm8r82r^8 (II)

Use equation (I) to write expression for the gravitational field at (2m,0m) by two-ball.

  g2=Gm2r22r^2 (III)

Use equation (I) to write expression for the gravitational field at (2m,0m) by four-ball.

  g4=Gm4r42r^4 (IV)

Use figure1, to calculate direction of field vector due to eight-ball.

  r^8=i^

Use figure1, to calculate direction of field vector due to two-ball.

  r^2=(cosθ)i^+sinθj^

Use figure1, to calculate direction of field vector due to four-ball.

  r^4=j^

Substitute (cosθ)i^+sinθj^ for r^2 in equation (III) to get g2.

  g2=Gm2r22((cosθ)i^+sinθj^) (V)

Write the expression for the total field at (2m,0m).

  gnet=g2+g8+g4 (VI)

Conclusion:

Substitute i^ for r^8, 2m for r8, 6.67×1011Nm2/kg2 and 0.150kg for in equation(II) to get g8.

  g8=6.67×1011Nm2/kg2(0.150kg)(2m)2(i^)=2.5×1012m/s2i^

Substitute (cosθ)i^+sinθj^ for r^2, 5m for r2, 6.67×1011Nm2/kg2, 27° for θ and 0.150kg for in equation(V) to get g2.

  g2=6.67×1011Nm2/kg2(0.150kg)(2m)2((cos27°)i^+sin27°j^)=2.0×1012((cos27°)i^+sin27°j^)=(1.8×1012i^+9.1×1013j^)m/s2

Substitute j^ for r^4, 1m for r4, 6.67×1011Nm2/kg2 and 0.150kg for in equation(IV) to get g4.

  g4=6.67×1011Nm2/kg2(0.150kg)(1m)2(j^)=1.0×1011m/s2j^

Substitute 2.5×1012m/s2i^ for g8, (1.8×1012i^+9.1×1013j^)m/s2 for g2 and 1.0×1011m/s2j^ for g4 in equation (VI) to get gnet.

  gnet=(1.8×1012i^+9.1×1013j^)m/s22.5×1012m/s2i^+1.0×1011m/s2j^=(4×1012i^+1×1011j^)m/s2

Therefore, the gravitational field at x=2cm, y=0 is (4×1012i^+1×1011j^)m/s2 .

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Chapter 7 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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