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IRISSAK [1]
3 years ago
6

If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D

when the force P

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer / Explanation

The question in the narrative is incomplete.

Kindly find the complete question below:

If the gap between C and the rigid wall at D is  initially 0.15 mm, determine the support reactions at A and  D when the force is applied. The assembly  is made of A36 steel

Procedure

Recalling the the equation of equilibrium and referencing the free body diagram of the assembly,

Therefore, ∑fₓ  = 0 ,

where, 20 ( 10³) - Fₐ - Fₙ = 0 --------------equation (1)

Now, recalling the compatibility equation, while utilizing the superposition method,

Therefore, δₓ - δfₓ

= 0.15  = 200(10³)(600) ÷ π/4 (0.05²)(200)(10⁹) - [ Fₐ (600) / π/4 (0.05²)(200)(10⁹) + Fₐ (600) π / 4 (0.05²)(200)(10⁹) ]

Solving this further,

We get: Fₐ = 20365.05 N

 Which is equivalent to = 20.4 kN.

Now, substituting the answer (Fₐ) into equation (1)

                Fₙ = 179634.95 N

                        = 180 kN

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a) Magnitude of force is 5.37N

b) Angle of force is - 37.88°

<h3>What is force?</h3>

In physics, we define a force is an influence that can change the motion of object. A force can cause an object with mass to change to accelerate.

Given force from above is 5.9N

∑Fₓ = F₁ₓ + F₂ₓ+F₃ₓ = 0

F₂ cos 44° + Fₓ = 0

Fₓ = - 5.9 cos44°

= - 4.24  N

Given force that acts downward as 7.4N

∑Fy= Fy1+ Fy2 +Fy3 =0

Fy+ 5.9 sin 44° - 7.4= 0

F =  3.30 N

F =√Fₓ² +Fy²

=  5.37 N

b)  angle of force= tan⁻¹ (3.30/-4.24)

= - 37.88°

To know more about force, refer

brainly.com/question/12970081

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A satellite is always being pulled by gravity.<br> a. True<br> b. False
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A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp
Anna71 [15]

\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

∴ The spring constant of the spring = 2500\text{ }\frac{kg}{s^{2}}

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3 years ago
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