Three balls A, B and C are kept in a straight line. The separation between A and C is 1 m, and B is placed at the midpoint between them. The masses of A, B, C are 100 g, 200 g and 300 g respectively. Find the net gravitational force on (a) A, (b) B, and (c) C
Answers
Answered by
451
A ------ 0.50 m -------- B -------- 0.50 -------- C
Ma = 0.100 kg Mb = 0.200 kg Mc = 0.300 kg
Net gravitational force Fa on A is towards right side.
Fa = G Ma Mb / 0.50² + G Ma Mc / 1.0²
= 6.67 * 10⁻¹¹ * 0.100 [ 0.200/ 0.50² + 0.300 /1 ] N
= 7.34 * 10⁻¹² N
Net gravitational force Fc on C is towards B, left side and its magnitude is equal to Fa ie., 7.34 * 10⁻¹² N
Net gravitational force on B is Fb & is towards C
Fb = G Mb [ Mc - Ma ]/ 0.50²
= 6.67 * 10⁻¹¹ * 0.200 * 0.200 * 4 = 1.07 * 10⁻¹¹ N
Ma = 0.100 kg Mb = 0.200 kg Mc = 0.300 kg
Net gravitational force Fa on A is towards right side.
Fa = G Ma Mb / 0.50² + G Ma Mc / 1.0²
= 6.67 * 10⁻¹¹ * 0.100 [ 0.200/ 0.50² + 0.300 /1 ] N
= 7.34 * 10⁻¹² N
Net gravitational force Fc on C is towards B, left side and its magnitude is equal to Fa ie., 7.34 * 10⁻¹² N
Net gravitational force on B is Fb & is towards C
Fb = G Mb [ Mc - Ma ]/ 0.50²
= 6.67 * 10⁻¹¹ * 0.200 * 0.200 * 4 = 1.07 * 10⁻¹¹ N
Answered by
49
Answer:
The correct answer is
a. 7.34 * 10^-12 N
b.1. 07*10^-11 N
c. 1.80 *10^-11 N
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