The system is released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied. If the mass of the cable and li
nks AB and BC can be neglected, and each pulley can be treated as a disk having a mass of 6 kg, determine the speed of the 10-kg block at the instant link AB has rotated θ = 90°. Note that point C moves along the vertical guide. Also, the cable does not slip on the pulleys.
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Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get: