Kinetics of rectilinear motion, relative motion and dependant motion :
Q. The 50 kg crate is projected along the floor with an initial speed of 7m/s at x = 0. The coefficient of kinetic friction µk = = 0.4. Calculate the time required for the crate to come to rest and corresponding distance x travelled. Solution : The free body diagram of the crate is as shown in in the fig.
Ʃ Fy = 0 (50 x g) = 0 ∴ N1 – (50
∴ N1 = 50 x 9.81 = 490.5 N Ʃ Fx = m x a ∴ - 0.4 N1 = m x a ∴ -0.4 x 490.5 = 50 x a ∴ a = - 3.924 m/s2 As per given , u = 7 m/s , v = 0
∴ v = u + at ∴ 0 = 7 + (-3.924) t ∴ t = 1.784 seconds Also ,
∴ v2 = u2 + 2 a s ∴ 02 = 72 + 2(-3.924)s ∴ s = 6.244 m Q. A body of mass ‘m’ is projected up at 25º inclined plane with an initial velocity velocit y of 15m/s. If the coefficient of friction µk = = 0.25. Determine the distance moved by the body up the plane and also find the time required to reach the highest point.( point.( technical pg 11.9) Solution : from the free body diagram as a s shown ;
Ʃ Fy = 0 ∴ N1 – mg mg cos25 = 0
∴ N1 = mg cos 25
Ʃ Fx = m x a ∴ – mg sin 25 – 0.25 N1 = m a ∴ – mg sin 25 – 0.25 mg cos 25 = m a ∴ a = -6.369 m/s2 u = 15 m/s
&
v=0
∴ v = u + at ∴ 0 = 15 + (-6.369) t ∴ t = 2.355 seconds ∴ v2 = u2 + 2 a s ∴ 0 = 152 + 2 (-6.369) s ∴ s = 17.66 m Q. The co-efficient of friction between blocks A and C and the horizontal surface are µ s = 0.24 and µk = 0.20. If Ma = 5.0kg and Mc = 10.0 kg, determine i) The tension in the chord.
ii) The acceleration of each block ( Refer Figure )
(technical pg 11.9) As per the figure aside and it’s F.B.D choose xA, xB, and xC . xA+2xB+ xC = l ;
aA + 2aB + aC = 0 ;
hence aB = - [aA+ aC]/2
let aA and aC be towards the origin i.e. towards the centre, then aB will be away from the origin i.e. downwards. The tensile forces are shown in the figure, the free body diagram be as shown , For A :
Ʃ Fy = 0
∴ NA = mA g = 5 x 9.81 = 49.05 N
Ʃ Fx = m x a
∴ T – 0.2 NA = 5 aA ∴ T – 0.2 (49.05 ) = 5 aA ∴ T – 9.81 = 5 aA
For C :
Ʃ Fy = 0 Ʃ Fx = m x a
---------------------------------
∴ NC = mA g = 10 x 9.81 = 98.1 N ∴ - T + 0.2 NC = -10 aC ∴ - T + 0.2 (98.1 ) = -10 aC
(i)
∴ - T + 19.62 = -10 aC
For C :
Ʃ Fy = m ay
----------------------------- (ii)
∴ 2T – 98.1 = 10 aB ∴ 2T – 98.1 = 10 (- [aA+ aC]/2 ) = - 5(aA+ aC ) ---------------------------- (iii)
Considering equation (i), (ii), (iii) and solving these equations we obtain;
∴ T = 33.63 N ∴ aA = 4.765 m/s 2 (→) ∴ aC = 1.401 m/s 2 (←) Substituting the values of aA & aC in aB
∴ aB = 3.083 m/s2 (↓)
Q. Block B rests on a smooth surface. If the co-efficient of static friction µs = 0.40 and kinetic friction = 0.30, determine the acceleration of e ach block if A is pushed by a force a) P = 30 kN
b) P = 250 N (Refer figure )
(technical pg 11.13)
Solution : Block A moves on block B if the external force P is greater than (Fr )max [ P > (Fr )max]
From the Free Body Diagram as shown Ʃ Fy = 0
N1-100=0 hence N1 = 100 N (Fr )max = µs* N1= 0.4*100= 40 N
Now if P < (Fr )max , then Block A does not move on the surface of block B, hence A and B will move together. Treating them as one object as shown in the figure. Ʃ Fx = m*a
i,e 30 = (250/9.81) * a, a = 1.772 m/s 2
But for P > (Fr )max as both the blocks move separately ( block A moves on B where as B moves on surface ) Considering F.B.D of both the blocks separately , As shown in the figure , For Block B ; Ʃ Fy = 0 N2 - N1 – 150 = 0 ,
∴ N1 = 100 N ∴ N2 = 250 N Ʃ Fx = m*a b
∴ 30 = (150/9.81) * a b ∴ ab = 1.962 m/s 2 For Block A ; Ʃ Fy = 0 N1 = 100 N Ʃ Fx = m*a 250 - µk N1 – (100/9.81) a b = (100/9.81) aA/B
∴ aA/B = 19.62 m/s 2 Also ; aA/B = aA – aB
∴ aA = 21.582 m/s2
MOTION OF A SET OF PARTICLES (Motion of connected rigid bodies) : Set of particles can be either connected set or particles or number of two or more unconnected particles moving in the same reference system. Thus the relationship between particles has to be taken into consideration. Let’s take the case of particle A and B as shown on diagram: Both particles are connected via inextensible cable carried over the pulleys. This imposes non penetrable condition between them: L
s A h 2 h sB
The additional length of the cable between the upper datum and the ceiling as well as the portion of the cable embracing the pulleys will remain constant during the motion thus does not play any role in the kinematical description.
Investigating mobility of the set of particles would define the number of independent coordinates that in our case is i 1 The path of a particle A is not identical with path of the particle B and the relation between them has to be described based on the joints involved. Thus except of the no-penetration condition the support at A has to be taken into consideration as well as the supports for the pulleys and body B. Having the basic condition of the inextensible length we can evaluate the relation between velocities of the particle A and B as a time derivative of the l . Thus
0 v A 2 v B
Then we can conclude that formation of the particle A in positive direction (away from the datum
v in the direction of s A) the particle B will move upward with velocity v B
Acceleration may be specified as a function of time, velocity, or position coordinate, or as a combined function of these. (a) Constant Acceleration At the beginning of the interval → t = 0, s = s0, v = v0 For a time interval t : integrating the following two equations
Substituting in the following equation and integrating will give the position coordinate:
Equations applicable for Constant Acceleration and for time interval 0 to t
(b) Acceleration given as a function of time, a = f(t) At the beginning of the interval
t
= 0,
s
= s0,
v
= v 0
For a time interval t : integrating the following equation :
Substituting in the following equation and integrating will give the position coordinate:
Alternatively, following second order differential equation may be solved to get the position coordinate:
(c) Acceleration given as a function of velocity, a = f(v) At the beginning of the interval → t = 0, s = s0, v = v0 For a time interval t : Substituting a and integrating the following equation :
Solve for v as a function of t and integrate the following equation to get the position coordinate:
Alternatively, substitute a = f (v) in the following equation and integrate to get the position coordinate :
(d) Acceleration given as a function of displacement, a = f(s) At the beginning of the interval
t = 0, s = s0, v = v0
For a time interval t : substituting a and integrating the following equation
Solve for v as a function of s : v = g ( s), substitute in the following equation and integrate to get the position coordinate:
It gives t as a function of s. Rearrange to obtain s as a function of t to get the position coordinate.
In all these cases, if integration is difficult, graphical, analytical, or computer methods can be utilized.
Motion with constant acceleration Motion with constant acceleration occurs in everyday life whenever an object is
dropped: the object moves downward with the constant acceleration
,
under the influence of gravity.
Fig. 8 shows the graphs of displacement versus time and velocity versus time for a body moving with constant acceleration. It can be seen that the displacement-time graph consists of a curved-line whose gradient (slope) is increasing in time. This line can be represented algebraically as (19)
Here,
is the displacement at time
: this quantity can be determined from
the graph as the intercept of the curved-line with the body's instantaneous velocity at time
.
-axis. Likewise,
is the
Figure 8: Graphs of displacement versus time and velocity versus time for a body moving with constant
acceleration
The velocity-time graph consists of a straight-line which can be represented algebraically as (20)
The quantity with the
is determined from the graph as the intercept of the straight-line
-axis. The quantity
is the constant acceleration: this can be determined
graphically as the gradient of the straight-line (i.e., the ratio Note that
, as shown).
, as expected.
Equations (19) and (20) can be rearranged to give the following set of three useful formulae which characterize motion with constant acceleration:
(21)
(22)
(23)
Here,
is the net distance traveled after
seconds.
Fig. 9 shows a displacement versus time graph for a slightly more complicated case of accelerated motion. The body in question accelerates to the right [since the gradient (slope) of the graph is increasing in time] between times and . The body then moves to the right (since is increasing in time) with a constant velocity (since the graph is a straight line) between times and . Finally, the body decelerates [since the gradient (slope) of the graph is decreasing in time] between times and .
Figure 9: Graph of displacement versus time
Using s-t curve, v-t & a-t curves can be plotted. Area under v-t curve during time dt = vdt == ds • Net disp from t1 to t2 = corresponding area under v-t curve
or s2 - s1 = (area under v-t curve) Area under a-t curve during time dt = adt == dv • Net change in vel from t1 to t2 = corresponding area under a-t curve
Or v2 - v1 = (area under a-t curve) 13
Two additional graphical relations: Area under a-s curve during disp ds= ads == vdv Net area under a-s curve betn position coordinates s1 and s2
Or ½ (v22 – v12) = (area under a-s curve) Slope of v-s curve at any point A = dv/ds Construct a normal AB to the curve at A. From similar triangles:
Velocity and position coordinate axes should have the same numerical scales so that the accln read on the x-axis in meters will represent the actual acceleration in m/s2
