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If a torque of M = 300 N⋅m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD t

Licemer1 [7]

Answer:

866.1 N

Explanation:

The torque on the flywheel = 300 N-m

The force from the hydraulic cylinder will generate a moment on CA about point A.

The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m

we know that moment = F x d

where F is the force, and

d is the perpendicular distance from the turning point = 1 m

Equating, we have

300 = F x 1

F = 300 N this is the frictional force that stops the flywheel

From F = μN

where F is the frictional force

μ is the coefficient of static friction = 0.4

N is the normal force from the hydraulic cylinder

substituting, we have

300 = 0.4 x N

N = 300/0.4 = 750 N

This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship

N = R sin (90 - 30)

750 = R sin 60°

750 = 0.866R

R = 750/0.866 = 866.1 N

3 0

3 years ago

It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less

solniwko [45]

Answer:

Acef

Explanation:

Edginuity 2021

5 0

3 years ago

Express the Internal Energy and Entropy as a Function of T and V for a homogeneous fluid. Develop the same relations using the i

DedPeter [7]

Answer:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

Explanation:

The internal energy is equal to:

$dU=C_{v} dT+(T(\frac{\delta P}{\delta T} )_{v} -P)dV$

The entropy is equal to:

$dS=C_{v} \frac{dT}{T} +(\frac{\delta P}{\delta T} )_{v} dV$

If we write the pressure derivative in terms of isothermal compresibility and volume expansivity, we have

$\frac{\delta P}{\delta T}=\frac{\beta }{\kappa }$

Replacing:

$dU=C_{v} dT+(T(\frac{\beta }{\kappa }) -P)dV$

$dS=C_{v} \frac{dT}{T} +(\frac{\beta }{\kappa } ) dV$

4 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".