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aleksandr82 [10.1K]
3 years ago
7

The arm of a crane at a construction site is 17.0 m long, and it makes an angle of 11.6 ◦ with the horizontal. Assume that the m

aximum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane withstand if the maximum load the crane can handle is 643 N? Answer in units of N · m.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

τ_max ≈ 10,708 N.m

Explanation:

To find max torque, we need the component of force that is perpendicular to the direction of vector r.

Now, Torque is the perpendicular component of the force times the distance.

Thus;

τ_max = F x r

τ_max = (643 cos 11.6°) (17 )

τ_max ≈ 10,708 N.m

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A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute
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          Mass, m = 75 g

         Velocity, v = 600 m/s

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          m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

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            m_{2} = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

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                                  = \frac{45}{50.075}

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               E = \frac{1}{2}mv^{2}

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             E' = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

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Therefore, energy lost will be calculated as follows.

           \Delta E = E  E'

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Thus, we can conclude that percentage n of the original system energy E is 99.9%.

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