If a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the
1 answer:
Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position
<h3>What is Displacement ?</h3>Displacement is the distance travelled in a specific direction. It is a vector quantity.
Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.
a) Let P be the Aster's final position from her initial position?
We can use bearing by using Cosine formula to solve this question.
P² = 70² + 82² - 2 × 70 × 82 cos 73
P² = 4900 + 6724 - 11480 cos 73
P² = 11624 - 3356.43
P² = 8267.57
P = √8267.57
P = 90.9 m
P = 90.9 - 28
P = 62.9 m
We can get the angle by using Sine rule
82/ sin Ф = 90.9 / sin 73
sin Ф = 0.8627
Ф = (0.8627)
Ф = 59.6°
Ф = 60°
b) She will head north west direction to return to her initial position
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Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows: