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Artist 52 [7]
3 years ago
10

The slotted arm OA rotates about a fixed axis through O. At the instant under consideration, θ = 36°, θ˙ = 36 deg/s, and θ¨ = 26

deg/s2. Determine the magnitude of the force F applied by arm OA and the magnitude of the force N applied by the sides of the slot to the 0.5-kg slider B. Neglect all friction, and let L = 0.72 m. The motion occurs in a vertical plane.
Physics
1 answer:
xz_007 [3.2K]3 years ago
4 0

Answer:

   F = 0.0545 N

Explanation:

Let's use Newton's second law for rotational movement

        τ = I α

The moment of inertia for a bar supported by some ends is

       I = 1/3 m L²

Torque is

            τ = F L

Let's replace

         F L = 1/3 m L² α

         F = 1/3 m L α

Let's reduce angular acceleration to SI units

       Alf = 26º / s² (π rad / 180º) = 0.454 rad / s²

Let's calculate

          F = 1/3  0.5  0.72  0.454

          F = 0.0545 N

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Explanation:

Acceleration a is expressed in the following formula:

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V_{o}=0 m/s  is the initial velocity of the projectile

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Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

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Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

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The work done is calculated from work-energy theorem as follows;

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h is height the total mass is raised = length of the rope = 70 ft

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