Figure 3
1 answer:
Answer:
3.92 N (straight down)
2.36 N (down, parallel to the plane)
Explanation:
The only force acting on the system is apparently that due to gravity. The mass of the system is the sum of the masses of the blocks, so is ...
0.20 kg + 0.20 kg = 0.40 kg
The force due to gravity is given by ...
F = ma = (0.40 kg)(9.8 m/s²) = 3.92 N . . . (straight down)
That force is acting straight down, so is at an angle with respect to the direction the blocks are constrained to move.
The force down the plane is (3.92 N)·sin(37°) ≈ 2.359 N . . . (down the plane)
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<em>Comment on the answer</em>
We have given two answers to the question, because in this frictionless system, the force acting normal to the plane of motion is irrelevant to the system dynamics. The question asked for the net force on the system, which is the force due to gravity, so we have given that magnitude also.
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Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
