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3 years ago

A load of 36 N attached to a spring hanging

Physics

1 answer:

givi [52]3 years ago

3 0

Answer:

The required force is 55.38 N.

Explanation:

First we have to find the spring constant K;

k = $\frac{mg}{d}$ = $\frac{36}{0.065}$ = 553.846 N/m

As,

F= -K $x$

or, F= - 553.846 × 0.10

or, F= -55.3846 N

Hence,

$F_{needed}$ = - $F_{spring}$

∴ $F_{needed}$ = 55.38 N

The required force to stretch it by this amount is 55.38 N.

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Read 2 more answers

Determine the components of reaction at the fixed support

denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are

$A_{(x)} = 400 \ N$ , $A_{(y)} = -500 \ N$ , $A_{(z)} = 600 \ N$ , $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ , $M_z = 0 \ N\cdot m$

Explanation:

Looking at the diagram uploaded we see that there are two forces acting along the x-axis on the fixed support

These force are 400 N and $A_{(x)}$ [ i.e the reactive force of 400 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(x)} - 400 = 0$

=> $A_{(x)} = 400 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the y-axis on the fixed support

These force are 500 N and $A_{(y)}$ [ i.e the force acting along the same direction with 500 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(y)} + 500 = 0$

=> $A_{(y)} = -500 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the z-axis on the fixed support

These force are 600 N and $A_{(z)}$ [ i.e the reactive force of 600 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(z)} - 600 = 0$

=> $A_{(z)} = 600 \ N$

Generally taking moment about A along the x-axis we have that

$\sum M_x = M_x - 500 (0.75 + 0.5) + 600 ( 1 ) = 0$

=> $M_x = 1225 \ N\cdot m$

Generally taking moment about A along the y-axis we have that

$\sum M_y = M_y - 400 (0.75 ) + 600 ( 0.75 ) = 0$

=> $M_y = 750 \ N\cdot m$

Generally taking moment about A along the z-axis we have that

$\sum M_z = M_z = 0$

=> $M_z = 0 \ N\cdot m$

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3 years ago

A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3

Inga [223]

$m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.$

The momentul of the system preserves:

$m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.$

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with $E$ .

Now, we write the energy conservation law:

$\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E$

From the above equation, you find $E$ , and then conclude that the sound energy can certainly not be greater than this.

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