Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward m
1 answer:
It sounds as though the two people are standing in front of the boat on opposite sides of it, so that they both make an angle of 30.0° with the axis of the boat, as in the attached free body diagram (ignoring the force of buoyancy and the weight of the boat).
By Newton's second law, the net vertical force is
∑ <em>F</em> = <em>P</em>₁ sin(60.0°) + <em>P</em>₂ sin(120.0°) - <em>R</em> = 0
where upward is positive and downward is negative, and the right side is 0 because the boat moves with constant velocity and thus zero acceleration.
We're told that <em>P</em>₁ = <em>P</em>₂ = 600 N, and we know sin(60°) = sin(120°), so the above reduces to
<em>R</em> = 2 <em>P</em> sin(60.0°) = 2 (600 N) sin(60.0°) ≈ 1040 N
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Answer:
Yes
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Answer:
4776.98 N is the minimum force to start the rise.
Explanation:
We can use the first Newton's law to find the minimum force to move the block.
So we will have:
Where:
- F is the force
- W(x) is the weight of the block in the x direction, W = mg*sin(15)
- F(f) is the static friction force (F(f) = μN), μ is the static friction coefficient 0.4.
Therefore 4776.98 N is the minimum force to move the block.
I hope it helps you!