2000, W. E. HaislerConservation of Linear Momentum - Part 21
Conservation of Linear Momentum
Selected Examples
Example 5-3: A mass of 50 kg is being lifted by a crane.
a)If the mass is to be accelerated at 1 m/sec2 upward, what force is required in the cable?
b)If the mass is to rise at a constant velocity of 5 m/s, what is the required force?
Solution
a) a = 1 m/sec2 in y direction; g (gravity) = -9.8 m/sec2 in y direction.
Step 1: Set-up coordinate system and draw system (free body).
Step 2: Write COLM equation (rate form) in table form.
Note: a = 1 m/sec2 in y direction; g (gravity) = -9.8 m/sec2 in y direction.
L.M. / / = / / + // + /
N / N / N / N
x / 50(0) / = / 0 / + / 50(0) / + / fx
y / 50(1) / = / 0 / + / 50(-9.8) / + / fy
z / 50(0) / = / 0 / + / 50(0) / + / fz
Step 3. Solve for unknown quantities.
fy = 50 - 50(-9.8) = 540 N; fx = fz = 0
b) Note: v = 5 m/sec in y direction (acceleration of mass is zero), g (gravity) = -9.8 m/sec2 in y direction. The Conservation of Linear Momentum table (rate form) is exactly the same except for the y component:
L.M. / / = / / + // + /
N / N / N / N
y / 50(0) / = / 0 / + / 50(-9.8) / + / fy
fy = 50(0)- 50(-9.8) = 490 N; fx = fz = 0
Design Problem: A mass of 900 kg is to be lifted as quickly as possible by the crane in Problem 5-3. If the cable has a tensile strength of of 5,000 lbf, what is the maximum acceleration that can be applied to the 900 kg mass without the cable breaking?
Example 5-4. Two cables hold a mass of 50 lbm between two
walls in the position shown. Determine the tensile force in each cable.
Step 1: Set-up coordinate system and draw system (free body).
We choose the free body (system) to be the juncture of the two cables and the cable supporting the mass:
Step 2: Write COLM equation (rate form) in table form.
LM / / = / / + // + / / + /
lbmft/sec2 / lbmft/sec2 / lbmft/sec2 / lbmft/sec2 / lbmft/sec2
x / 50(0) / = / 0 / + / 50(0) / + / -f1x / + / f2x
y / 50(0) / = / 0 / + / 50(-32.174) / + / f1y / + / f2y
z / 50(0) / = / 0 / + / 50(0) / + / f1z / + / f2z
Note that we have 6 unknown force components. However, the x, y and z components of and are related by geometry through the angle each cable makes with the horizontal.
For simplicity, we note that this is closed system (static solid) and we reduce the table by eliminating the rate terms. When we recognize a problem to be a static equilibrium problem, we can always make this simplifying step.
LM / 0 / = // + / / + /
lbmft/sec2 / lbmft/sec2 / lbmft/sec2
x / 0 / = / 50(0) / + / -f1cos(25) / + / f2cos(60)
y / 0 / = / 50(-32.174) / + / f1sin(25) / + / f2sin(60)
z / 0 / = / 50(0) / + / f1cos(90) / + / f2cos(90)
Using direction cosines to obtain the x, y and z components of each force vector, the last result can be written as (see notes for chapter 5):
LM / 0 / = // + / / + /
lbmft/sec2 / lbmft/sec2 / lbmft/sec2
x / 0 / = / 50(0) / + / f1cos(155) / + / f2cos(60)
y / 0 / = / 50(-32.174) / + / f1cos(65) / + / f2cos(-30)
z / 0 / = / 50(0) / + / f1cos(90) / + / f2cos(90)
Evaluating the trigonometric functions yields:
LM / 0 / = // + / / + /
lbmft/sec2 / lbmft/sec2 / lbmft/sec2
x / 0 / = / 50(0) / + / -f1(0.906) / + / f2(0.5)
y / 0 / = / 50(-32.174) / + / f1(0.423) / + / f2(0.866)
z / 0 / = / 50(0) / + / f1(0) / + / f2(0)
Step 3. Solve for unknown quantities.
Consequently from y and z equilibrium equations, we have two equations and two unknowns:
Solving this system of equations, we obtain
Convert from
Example 5-2. Consider the collision of two balls in the absence of any gravitational field. Figure (a) shows the balls before collision. The mass of ball 1 is 2.5 kg and the mass of ball 2 is 5 kg. The velocity vector of each ball prior to collision are given by: and (m/s).
a) If the velocity vector of ball 2 after collision is (m/s), determine the velocity vector of ball 1 after collision.
b) If the velocity vector of ball 2 after collision is the same as above and the velocity vector of ball 1 after collision is (m/s), what would be the external force which must have acted on the balls during collision if the time period for the collision was 0.1 s (i.e., what impulse has to be applied).
For this problem, we assume that the system is the two balls and the balls remain in the system after collision. We use the integrate form (for a finite time period) of COLM:
or, for a finite time period :
or,
where
If the external force is constant, the last term (which is the impulse) can be replaced by .
a) Writing out the (x,y,z) components of COLM in table form gives the following. Note that since the system is closed (no mass enters or leaves the system), the linear momentum entering and leaving is zero. In addition, no external forces are applied to the system.
/ - / / = / / + // + / / /
kg m/s / kg m/s / kg m/s / kg m/s / kg m/s / kg m/s
x / / + / 5(1) / - / 2.5(4) / - / 5(-2) / = / 0 / + / 0
y / / + / 5(-2) / - / 2.5(7) / - / 5(1) / = / 0 / + / 0
z / / + / 5(4) / - / 2.5(5) / - / 5(3) / = / 0 / + / 0
Solving these three equations gives: , and . Thus,
b) Since the force is constant with time, we replace by
. Recall that and (m/s), and that an impulse is applied for 0.1 s. Writing out the (x,y,z) components of COLM in table form gives the following.
/ - / / = / / + // + / / /
kg m/s / kg m/s / kg m/s / kg m/s / kg m/s / kg m/s
x / 2.5(-1) / + / 5(1) / - / 2.5(4) / - / 5(-2) / = / 0 / + /
y / 2.5(12) / + / 5(-2) / - / 2.5(7) / - / 5(1) / = / 0 / + /
z / 2.5(2) / + / 5(4) / - / 2.5(5) / - / 5(3) / = / 0 / + /
Substituting , solving for the force components and writing the result in vector form gives: .
Example 5-5: Determine the general expression in a vertical column of a static fluid. We consider a fee body (system) to be a differential cube of size x by y by z in a Cartesian coordinate system and located at x,y,z. The cube is acted upon by the fluid pressure P(x,y,z) and gravity g in the -z direction. We now write the COLM table. We assume that system is steady state so that all time rates are zero.
LM / / = / / + // + /
N / N / N / N
x / m (0) / = / 0 / + / m (0) / + / [P(x)-P(x+x)] yz
y / m.(0) / = / 0 / + / m (0) / + / [P(y)-P(y+y)] xz
z / M (0) / = / 0 / + / m (-g) / + / [P(z)-P(z+z)] xy
From the x LM equation: divide by xyz to obtain . Take the lim x0 to obtain . Similarly for the y LM equation: . These latter two equations imply that P is not a function of x or y. For the z LM equation, we obtain
In the limit as x, y, z 0, the term on the left becomes and the mass per unit volume becomes the mass density: . Note that =(x,y,z). Hence, we have
Suppose we consider a column of water and integrate the above from z1 to z2 . We obtain
Example 5-7: Fluid is moving through a curved pipe. Assume steady flow (no acceleration). We take as given the linear momentum rate into and out of the pipe to be: and . Determine the force f required on the boundary.
Since the flow is assumed to be steady state, then . We assume the system is the boundary of the pipe. We write the COLM equation (rate equation) in table:
LM / / = / / + // + /
lbmft/sec2 / lbmft/sec2 / lbmft/sec2 / lbmft/sec2
x / msys(0) / = / 25 / - / 15 / + / msys(0) / + / fx
y / msys(0) / = / 0 / - / 20 / + / msys(-g) / + / fy
z / msys(0) / = / 0 / - / 0 / + / msys(0) / + / fz
From the table: or, in vector form, (lbmft/sec2). Convert to lbf (divide by gc) to obtain (lbf).