System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 4, Problem 4.50P
To determine

(a)

The equation of motion for the system.

Expert Solution
Check Mark

Answer to Problem 4.50P

The equation of motion for the system is 8.33v˙+v=34.51.

Explanation of Solution

Given information:

The mass of the block on an inclined plane is 50kg, the angle of inclination is 25°, the friction coefficient between the block and the plane is 6Ns/m.

Figure-(1) shows the mechanical system diagram of the block on the inclined plane condition.

System Dynamics, Chapter 4, Problem 4.50P

Figure-(1)

Here, the component of force acting in the horizontal direction along the inclined plane is mgsinθ, the component of force acting in vertical direction and opposite to reaction force is mgcosθ.

Write the expression for equation of motion for the system along the inclined plane.

mv˙=mgsinθcv ..... (I)

Here, the force acting on the block is mv˙, the velocity of the block is v ,the acceleration of the block is v˙, the mass of the block is m, the angle of inclination is θ, the coefficient of friction is c, acceleration due to gravity is g.

Calculation:

Substitute 50Kg for m, 25° for θ, 9.81m/sec2 for g, and 6Ns/m for c in Equation (I)

(50kg)v˙=(50kg)(9.81m/sec2)sin(25°)(6Ns/m)v(50kg)v˙+(6Ns/m)v=(490kgm/sec2)sin(25°)(50kg)v˙+(6Ns/m)v=207.08kgm/sec28.33v˙+v=34.51.

Conclusion:

The equation of motion for the system is 8.33v˙+v=34.51.

To determine

(b)

The equation of motion for speed V(t) if the initial speed of the block is 5m/s.

Expert Solution
Check Mark

Answer to Problem 4.50P

The equation of motion for speed V(t) is V(t)=34.54834.548et8.33.

Explanation of Solution

Given information:

The initial speed of the block is 5m/s.

Write the expression for equation of motion for the system.

8.33v˙+v=34.548 ..... (IV)

Apply Laplace transform for Equation (IV)

8.33sV(s)+V(s)=34.548s(8.33s+1)V(s)=34.548ss(8.33s+1)V(s)=34.548V(s)=34.548s(8.33s+1) ...... (V)

Apply partial fraction technique for Equation (V).

As+B8.33s+1=34.548s(8.33s+1)A(8.33s+1)+Bss(8.33s+1)=34.548s(8.33s+1)A(8.33s+1)+Bs=34.548 ..... (VI)

Substitute 0 for s in equation (VI)

A(8.33(0)+1)+B(0)=34.548A+B(0)=34.548A=34.548

Equate coefficient on both sides of Equation

0=8.33A+B ...... (VII)

Substitute 34.548 for A in Equation (VII)

0=8.33(34.548)+BB=287.784.

Calculation:

Substitute 34.548 for A and 287.784 for B in Equation (VI)

34.548(8.33s+1)+(287.784)ss(8.33s+1)=34.548s(8.33s+1)34.548(8.33s+1)s(8.33s+1)(287.784)ss(8.33s+1)=34.548s(8.33s+1)

Substitute 34.548(8.33s+1)s(8.33s+1)(287.784)ss(8.33s+1) for 34.548s(8.33s+1) in Equation (V)

V(s)=34.548(8.33s+1)s(8.33s+1)(287.784)ss(8.33s+1)V(s)=34.548s34.548s+18.33

Take inverse Laplace transform.

V(t)=34.54834.548et8.33.

Conclusion:

The equation of motion for speed V(t) is V(t)=34.54834.548et8.33.

To determine

(c)

The steady-state speed of the block

The time required to reach the speed.

Expert Solution
Check Mark

Answer to Problem 4.50P

The steady-state speed of the block is 34.548m/s

The time required to reach the speed 33.32s.

Explanation of Solution

Write the expression for first order differential equation along with dynamic value.

τv˙+v=Ku ...... (VIII)

Here, the time constant is τ, the steady-state speed is K, the velocity of the block is v.

Write the expression for time required to reach the steady-state speed

t=4τ ...... (IX)

Here, the time required to reach the steady-state is t.

Calculation:

Compare Equation (IV) and Equation (VIII).

τ=8.33

Substitute 0 for v˙ in Equation (IV)

8.33(0)+v=34.548v=34.548

Substitute 8.33 for τ in Equation (IX)

t=4(8.33)=33.32s.

Conclusion:

The steady-state speed of the block is 34.548m/s

The time required to reach the speed 33.32s.

To determine

(d)

The effect of initial velocity on the steady-state speed.

Expert Solution
Check Mark

Explanation of Solution

Write the expression for steady-state speed.

v˙=0

Since the steady-state speed is zero therefore, the initial velocity is independent of the steady-state speed and hence, the initial velocity does not affect the steady-state speed value.

Conclusion:

The initial velocity does not affect the steady-state speed value.

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Chapter 4 Solutions

System Dynamics

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