Problem 3: When x = 10 ft, the crate has a speed of 20 ft/s which is increasing at 6 ft/s². Determine the direction of (i.e. the angle of) the crate's velocity and the magnitude of the crate's acceleration at this instant. Also find the horizontal components of velocity and acceleration at this instant (by projecting the velocity and acceleration vectors along the x-axis). Ans: 0 = 39.81°, |a| = 16.3 ft/s², vx = -15.36 ft/s, ax = -14.28 ft/s² (F12-30)
Examine pls
At x = 10 ft, the position of the center of the crate to the vertical axis is given by
y = (1/24)x2, so y = (1/24)102 = 4.17 ft.
The velocity of the crate at this instant is 20 ft/s, increasing at 6 ft/s^2, so the acceleration of the crate is 6 ft/s^2.
To determine the direction of the velocity, we can use the tangent of the angle between the velocity vector and the positive x-axis.
The tangent of the angle is equal to the vertical component of the velocity divided by the horizontal component of the velocity.
The horizontal component of the velocity is given by vx = vcos(θ), and
the vertical component of the velocity is given by vy = vsin(θ),
where v is the magnitude of the velocity and θ is the angle between the velocity vector and the positive x-axis.
Since the crate is moving downwards, the vertical component of the velocity is negative, so vy = -20sin(θ).
The horizontal component of the velocity is positive, so vx = 20cos(θ).
We can use the tangent of the angle to find the angle itself:
tan(θ) = vy/vx = -20 sin(θ)/20 cos(θ)
Solving for θ, we find that
θ = tan-1(-20sin(θ) / 20 cos(θ))
The inverse tangent function returns an angle in the range (-90, 90) degrees.
Using a calculator, we find that the angle is approximately 39.81 degrees, measured counterclockwise from the positive x-axis.
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