The following coplanar forces pull on a ring 200N at 30 degrees and 500N at 80 degrees and 300N at 240 degrees and an unknown fo
1 answer:
MAGNITUDE AND DIRECTION OF A VECTOR
Given a position vector →v=⟨a,b⟩,the magnitude is located by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba)
<h3>What is the formula of magnitude?</h3>
The formula to determine the extent of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the procedure to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)
To learn more about magnitude and direction, refer
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Answer:
The maximum acceleration of the system is 359.970 centimeters per square second.
Explanation:
The motion of the mass-spring system is represented by the following formula:
Where:
- Position of the mass with respect to the equilibrium position, measured in centimeters.
- Amplitude of the mass-spring system, measured in centimeters.
- Angular frequency, measured in radians per second.
- Time, measured in seconds.
- Phase, measured in radians.
The acceleration experimented by the mass is obtained by deriving the position equation twice:
Where the maximum acceleration of the system is represented by .
The natural frequency of the mass-spring system is:
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
If and
, the natural frequency is:
Lastly, the maximum acceleration of the system is:
The maximum acceleration of the system is 359.970 centimeters per square second.
The image is missing, so i have attached it;
Answer:
A) V_rel = [-(2.711)i - (0.2588)j] m/s
B) a_rel = (0.8637i + 0.0642j) m/s²
Explanation:
We are given;
the cruise ship velocity; V_b = 1 m/s
Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s
Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²
Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)
Thus,
V_a = V_b + (ω•r_ba) + V_rel
Now, V_a = 0. Thus;
0 = V_b + (ω•r_ba) + V_rel
V_rel = -V_b - (ω•r_ba)
From the image and plugging in relevant values, we have;
V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)
V_rel = - (cos15)i - (sin15)j - 1.745i
Note that; k x j = - i
V_rel = [-(2.711)i - (0.2588)j] m/s
B) Let's write the acceleration at A with respect to B in terms of a_b.
Thus,
a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel
a_a and a_b = 0.
Thus;
0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel
a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)
Plugging in the relevant values with their respective position vectors, we have;
a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])
a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i
Note that; k x j = - i and k x i = j
Thus,simplifying further ;
a_rel = 0.8637i + 0.0642j m/s²
