Answer:
the time of motion of the ball is 6.89 ms.
Explanation:
Given;
angular speed, ω = 38 rad/s
angular distance, θ = 15 degrees
Angular distance in radian;
Time of motion is calculated as;
Therefore, the time of motion of the ball is 6.89 ms.
The required work to hold the crate above the ground is 150 joule.
We need to know about the work done to solve this problem. The work done by an object depends on the force applied and the distance. The work is proportional to force and displacement. It can be written as
W = F . s
where W is work done, F is the force and s is displacement
From the question above, the given parameters are
F = 100 N
s = 1.5
Thus, the required work to hold the crate above the ground can be calculated
W = F . s
W = 100 . 1.5
W = 150 joule
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Below is the solution:
<span>T2cos(30) - T1cos(50) = 0
</span><span>T1sin(50) + T2sin(30) - (75 lbs.)*(accel. grav.) = 0
</span><span>T2cos(30) - T1cos(50) = 0 --> T1 = T2cos(30)/cos(50)
</span>
<span>T1sin(50) + T2sin(30) - (75 lbs.)*(accel. grav.) = 0
</span>(<span>T2cos(30)/cos(50))sin(50) + T2sin(30) - (75 lbs.)*(accel. grav.) = 0 --> Solve for T2
</span><span>T1 = -T1cos(50)i + T1sin(50)j
T2 = T2cos(30)i + T2sin(3)j
</span>
<span>(T2cos(30)/cos(50))sin(50) + T2sin(30) - (75 lbs.)*(accel. grav.) = 0 -->
T2[(cos(30)/cos(50))sin(50) + sin(30)] = 75*(grav) -->
T2 = 75*grav/ [(cos(30)/cos(50))sin(50) + sin(30)]
</span>
<span> T2 = 1566.49 </span>
