December 2018 Solutions

Q. 1 Solve any four [5 X 4]
a) Find the resultant of the parallel force system shown in Figure 1 and locate the same with respect to point C.[05 M]

Ans :

Net Force along x-axis

\sum F_y=0\;\rightarrow\left[Left\;to\;right\;Positive\right]

Net force along y-axis

Consider downward force negative and upward force positive

\sum F_y=15-60+10-25

\sum F_y=-60\;N

Resultant Force = $\sqrt{Fx^2+Fy^2}$

=\sqrt{0+\left(-60\right)^2}

=\sqrt{60^2}

∴ The Resultant Force (R) = 60 N

Now, $\tan\left(\theta\right)=\frac{\sum F_y}{\sum F_x}$

\theta=\tan^{-1}\left(\frac{\sum F_y}{\sum F_x}\right)

\theta=\tan^{-1}\left(\frac{-60}0\right)

\theta=-90

By Varignon’s Theorem,

\sum M_C^F=R\times d

Where, d=The distance of the resultant force from point C.

Assuming ‘R’ to be on the right of point C

\left(15\times40\right)-\left(10\times30\right)+\left(25\times80\right)=60\times d

600-300+2000=60d

2300=60d

d=\frac{2300}{60}

d=38.33 m

The resultant force is 60 N downward. It is located 38.33 mtr. away to the right of point C.

Q. 1 b) Using Instantaneous Centre of rotation (ICR) method, find the velocity of point A for the instant shown in figure 2. Collar B moves along the vertical rod, whereas link AB moves along the plane which is inclined at $25˚$ .[05 M]

Ans :

Given, The collar B moves along the vertical rod, whereas link AB moves along the plane which is inclined at $25˚$ .

The $\theta=45˚$

By using Sine rule,

\frac{AB}{\sin\left(I\right)}=\frac{BI}{\sin\left(A\right)}=\frac{AI}{\sin\left(B\right)}

BI=\frac{1.2\times\sin\left(60\right)}{\sin\left(75\right)}=1.076m

AI=\frac{1.2\times\sin\left(45\right)}{\sin\left(75\right)}=0.878m

W_{AB}=BI\times V_{\boldsymbol B}\rightarrow eq.1

Putting the values in equation 1

W_{AB}=1.076\times1.5

W_{AB}=1.614\;rad/s

Also, $W_{AB}=AI\times V_A$

V_A=\frac{W_{AB}}{AI}

V_A=\frac{1.614}{0.878}

V_A=1.838\;m/s

The velocity of point A for the given instance is 1.838 m/s.

Q.1 c) If the support reaction at A, for the beam shown in figure 3, is zero, then find force ‘P’ and the support reaction at B.[05 M]

Ans :

Net force along the x-axis

$\sum F_x=0\;(\rightarrow)$ [Left to right positive]

Net force along y-axis

Consider, downward force negative and upward positive

\sum F_y=10-60+R-P=0;

Where R $\rightarrow$ Reaction,

P $\rightarrow$ Force.

R-P=50\rightarrow equation\;1

\sum M_B^F=(P\times2)-(60\times1.667)-(10\times7)=0

$P=85.01\rightarrow$ equation 2

Putting the value of ‘P’ in the equation 1

R-85.01=50

R=85.01+50

R=135.01\;kN

The magnitude of force P is 85.01 kN and the reaction is 135.01 kN.

Q. 1 d) From the top of a tower, 28 m high, a stone is thrown vertically up with a velocity of 9 m/s. After how much time will the stone reach the ground ? With what velocity does it strike the ground ? [05 M]

Ans :

The given parameters,

The final velocity (V) = o m/sec

The accelerometer due to gravity (g) = 9.812 m/s².

From A to B ;

The Initial Velocity (u) = 9 m/sec

The Final Velocity (v) = 0 m/sec

acc = gravity = -9.812 m/s²

[downwards to upwards considered negative]

Using the 1^st equation of motion

v = u + at,

0 = 9 + (-9.812 x t₁)

t₁ = 0.917 seconds

To find the distance travelled ‘x’ we use the 3^rd equation

v²=u²+2(-9)x

0²= 9²+2(-9.128)x

x=9.128mtr.

From B to C

Initial velocity (u) = o m/sec

The acceleration due to gravity (g) = 9.812 m/s² .

The distance travelled (s)=x+28=4.128+28+32.128 mtr.

To find the final velocity using the 3^rdequation of motion

v²=u²+2gs

v²=0+2×9.812×32.128

\text{v=}\sqrt{2\times9.812\times32.128}\;=25.109m/sec

To find the time (t₂) using the 1^st equation of motion

v=u+9t₂

v=0+9t₂

t_2=\frac vg=\frac{25.109}{9.812}=2.559\;sec

Total time (T) = t₁+t₂

= 0.917+2.559

=3.476 sec.

The stone will strike the ground after 3.476 sec. at a velocity of 25.109 m/sec.

Q. 1 e) For the truss shown in figure 4, find: (i) zero force members, if any (Justify your answer with FBD), (ii)support reactions at C and D.[05 M]

Ans :

The member F_C is a zero force member as there is no F external load on F and there are two other collinear E members.

Net force in horizontal direction $\sum F_x=0$

H_D=0

Net force in vertical direction $\sum F_y=0$

i ) -20-50+R_C+V_D=0

[Force in upward direction positive and downward direction is negative]

ii ) L/2 x L/2 x L/2 R_C+R_D=70 $\rightarrow$ equation 1

\sum M_C^F=0

+(R_{Dx}.L)-\left(20\times\frac L2\right)+\left(50\times\frac L2\right)=0

R_D=85kN

Zero force members-F_C

The magnitude of the support reactions at C and D are, H_D=0, R_D=-15 kN and R_C=85kN respectively.

Q. 2 a) For the composite lamina shown in figure 5, determine the coordinates of its centroid. [08 M]

Ans :

The area of the shaded region =The rectangle ABFG + rectangle OCDF + Quarter Circle OCB – triangle AEG.

For Rectangle ABFG

Area (A_i)=90×60=5400 cm²

x co-ordinate (x_i)= $\frac{-60}2=-30$

y co-ordinate (y_i)= $50-\frac{90}2=5$

\therefore A_iX_i=540\times(-30)=-162000

\therefore A_iy_i=540\times5=27000

For Rectangle OCDF

Area (A_i)=40×50=2000 cm²

x co-ordinate (x_i)= $\frac{50}2=25$

y co-ordinate (y_i)= $\frac{40}2=-20$

\therefore A_iX_i=2000\times(25)=50000

\therefore A_iy_i=2000\times-20=-40000

For Quarter circle OCB

Area (A_i) = $\frac14\times\mathrm\pi\times\mathrm r^2$

= $\frac14\times\mathrm\pi\times\mathrm 50^2$

=1963.495 cm²

x co-ordinate (x_i) = 21.22

y co-ordinate (y_i) =21.22

\therefore A_iX_i=1963.495\times(21.22)=41665.3639

\therefore A_iy_i=1963.495\times21.22=41665.3639

For Triangle AEG

Area (A_i) = $-\frac12\times75\times90=-3375\;cm^2$

x co-ordinate (x_i) = -35

y co-ordinate (y_i) =-10

\therefore A_iX_i=-3375\times(-35)=113125

\therefore A_iy_i=3375\times-10=33750

\therefore\sum A_i=5400+2000+1963.495+3375=5988.4950\;cm^2

\therefore\sum A_ix_i=-162000+50000+41665.3639+118125=47790.3639

\therefore\sum A_iy_i=27000-40000+41663.3639+33750=62415.3639

\overline x\;=\;\frac{\sum A_ix_i}{\sum A_i}=\frac{47790.3639}{5988.495}=7.98

\overline y\;=\;\frac{\sum A_iy_i}{\sum A_i}=\frac{62415.3639}{5988.495}=10.425\;cm^2

The co-ordinates are (7.98, 10.425)

Q. 2 b) Replace the force system shown in figure 6 with a single force and couple system acting at point B. [05 M]

Ans :

In $\triangle$ AGD,

\tan\left(b\right)=\frac{height}{base}=\frac{GD}{DA}

b=\tan^{-1}\left(\frac34\right)

b=36.87˚

In $\triangle$ AEF,

\tan\left(a\right)=\frac{AE}{AF}

a=\tan^{-1}\left(\frac{AE}{AF}\right)

a=\tan^{-1}\left(\frac32\right)

a=56.32˚

In $\triangle$ FHD,

c=\tan^{-1}\left(\frac{DC}{CH}\right)

c=\tan^{-1}\left(\frac62\right)

c=71.57˚

Net horizontal force,

\sum F_x=1000\cos\left(b\right)+632\cos\left(c\right)-722\cos\left(a\right)

\sum F_x=1000\cos\left(36.87\right)+632\cos\left(71.57\right)-722\cos\left(56.32\right)

=599.42

Net vertical force,

\sum F_y=-100\sin\left(b\right)+632\sin\left(c\right)-722\sin\left(a\right)

\sum F_y=-100\sin\left(36.87\right)+632\sin\left(71.57\right)-722\sin\left(56.32\right)

=-601.23

\therefore The\;resul\tan t\;force=\sqrt{F_x^2+F_y^2}

R=\sqrt{599.42^2+{(-601.23)}^2}

R=843.99N

\theta=\tan^{-1}\left(\frac{\sum F_y}{\sum F_x}\right)=\tan^{-1}\left(\frac{-601.23}{599.42}\right)=-45.086˚

\sum M_B^F=-\left[722\cos\left(56.32\right)\times3\right]+\left[1000\cos\left(36.87\right)\times6\right]-\left[632\sin\left(71.57\right)\times2\right]

\sum M_B^F=2399.66Nm

The magnitude of the resultant force and couple at B are 848.99 N and 2399.66 Nm clockwise.

Q. 2 c) The link CD of the mechanism shown in figure 7 is rotating in counterclockwise direction at an angular velocity of 5 rad/s. For the given instance, determine the angular velocity of link AB [07 M]

Ans :

In $\triangle$ EBC,

Using the Sine rule,

\frac{EB}{\sin\left(75\right)˚}-\frac{BC}{\sin\left(45\right)˚}-\frac{CE}{\sin\left(60\right)˚}

EB=\frac{200\times\sin\left(75\right)˚}{\sin\left(45\right)˚}=0.27\;m

EB=\frac{200\times\sin\left(60\right)˚}{\sin\left(45\right)˚}=0.24\;m

W_CD=V_c x CD

Angular velocity=(V_c )= $\frac5{0.1}=50\;m/sec$

W_BC= V_c x CE = 50 x 0.24 = 12 rad/sec

W_BC= V_B x EB

V_B=\frac{12}{0.27}=44.44\;m/sec

W_AB= V_B x AB=44.44 x 0.15 = 6.666 rad/sec

The angular velocity of link AB is 6.666 rad/sec

Q. 3 a) Cylinder A (diameter 1 m, Weight 20 kN) and cylinder B (diameter 1.5 m, weight 40 kN ) are arranged as shown in Figure 8. Find the reactions at all the contact points. All contacts are smooth. [06 M]

Ans :

In $\triangle$ DAE and $\triangle$ DAF

DA = DA [Common side]

∠ DEA = ∠ DFA = 90°

By RHS rule

∆ DAE ≅ ∆ DAF [Congruent to each other]

∠ DEA = ∠ DFA

∠ DAE + ∠ DAF = 2 x ∠DAE

= 180-60

= 120°

∴ ∠ DAE = ∠ DAF = $\frac{120˚}2$

∴ ∠ DAE = ∠ DAF = 60°

In ∆ DAF,

\tan\left(60\right)˚=\frac{DF}{AF}

AF\;=\;\frac{0.75}{\tan\left(60\right)}\left[\because\;Diameter\;B\;\left(\varnothing\right)\;=\;1.5\;m\right]

AF = 0.433 mtr.

Now, AG = AF+FC+CG

1.5 = 0.433+ DB + HI

1.5 – 0.433 – HI = DB

1.5 – 0.433 – 0.5 = DB [∵ Diameter of A(∅)=1.0 m]

DB = 0.567 mtr.

In ∆ DBH,

\cos\left(D\right)=\frac{DB}{DH}=\frac{0.567}{0.75+0.5}=0.4536

D=\cos^{-1}\left(0.4536\right)=63.025˚

Considering Cylinder A :

Using Lami’s Theorem

\frac{20}{\sin\left(180-63.025\right)}=\frac A{\sin\left(90\right)}=\frac B{\sin\left(90+63.025\right)}

A = 22.44 kN ; B = 10.1795 kN

Considering Cylinder B :

\sum F_x=0\;;\;300\;D\;\cos\left(30\right)-A\;\cos\left(63.025\right)=\;0\;

D=11.75 N

\sum F_y=0\;;\;C\;-\;40\;+\;D\;\sin\left(30\right)-A\;\sin\left(63.025\right)=\;0\;

C = 25.876 kN ; A = 22.44 kN ; B = 10.1795 kN ; D = 11.75 kN.

Q. 3 b) Using Principle of Virtual Work, determine the force P which will keep the weightless bar AB in equilibrium. Take length AB as 2 m and length AC as 8 m. The bar makes an angle of 30° with horizontal. All the surface in contact are smooth. Refer Figure 9. [06 M]

Ans :

For active force (-P) [force from right to left]

Co-ordinate of point of application 8 Cos (30°)

Virtual displacement -8 Sin (30) dθ

For active force -800 [force from top to bottom]

Co-ordinate of point of application 6 Sin (30°)

Virtual displacement 6 Cos (30°) dθ

By the principle of virtual work

∑ W=0

-P(-8 Sin(30°) dθ )-800 (6 Cos (30°) dθ)=0

P=1039.23 N

The magnitude of the force P is 1039.23 N.

Q. 3 c) Velocity-time diagram for a particle travelling along a straight line is shown in Figure 10. Draw acceleration-time and displacement-time diagram for the particle. Also find important values of acceleration and displacement.[08 M]

Ans :

From 0-5 Seconds

Initial velocity (u)=0 m/sec

Final velocity (v)-at

a= $\frac vt=\frac{20}5=4\;m/sec^2$

S=ut+\frac12at^2

S=\frac12\;at^{2\;}\left[\because u=0\right]

S=\frac12\;\times4\times5^2=50\;mtr.

From 5-20 seconds

Initial velocity = Final velocity u=v=20 m/sec

a=ut

S=20 x 15

S = 300 mtr.

Total displacement after 20 seconds

S₀ – 5 + S_{5 to 20}= 300 + 50 = 350 mtr

From 20-30 seconds

Initial Velocity (u) = 20 m/sec

Final velocity (v) = 60 m/sec

v = u + at

a\;=\;\frac{v\;-\;u\;}t=\frac{60-20}{10}=4\;m/sec^2

a\;=\;\left(20\times10\right)+\frac12\times4\times10^2

S=400 mtr.

Total displacement after 30 seconds = S_{0 to 20} + S_{20 to 30}

= 350 + 400

= 750 mtr

Q. 4 a) Find the force ‘F’ to have motion of block A impending up the Plane. Take coefficient of friction for all the surfaces in contact as 0.2. Consider the wedge B as weightless. Refer Figure 11.[07 M]

Ans :

Consider block A

∑ F_x=0

-B Cos (15°)- S Cos (75°) + A Cos (75°) + R Cos (15°) = 0

0.2 S Cos (15°) – S Cos (75°) + 0.2 R Cos (75°) + R Cos (15°) = 0

S (-0.2 Cos(15°) – Cos (75°))+ R (0.2 Cos (75°) + Cos(15°)) = 0 → ①

∑ Fy=0

-200-B Sin (15°)- S Sin (75°) + R Sin (15°) – A Sin (75°) = 0

-0.2 S Sin (15°) – S Sin (75°) – 0.2 R Sin (75°) + R Sin (15°) = 200

S(-0.2 Sin (15°) + Sin (75°)) + R(-0.2 Sin (75°) + Sin (15°)) = 200 → ②

Now solving ① and ②

S = 212.019 N

R = 94.168 N

Consider Block Wedge B,

∑ F_x=0

-F + S Cos (75°) + Sin (75°) + D = 0

F = S Cos (75°) + 0.2 S Sin (75°) + 0.27

∑ Fy=0

T – S Sin (75°) + Cos (75°) = 0

T = S Sin (75°) – 0.2 S Cos (75°)

T = 193.3197 N

F = S Cos (75°) + 0.2 S Sin (75°) + 0.2 TN

F = 134.597 N

The magnitude of the force F is 134.597 N.

Q. 4 b) Three forces F₁, F₂ and F₃ act at the origin of Cartesian coordinate axes system. The force F₁ (= 70 N) acts along OA whereas F₂(=80 N) acts along OB and F₃ (= 100 N) acts OC. The co-ordinates of the points A, B and C are (2, 1, 3), (-1, 2, 0) and (4, -1, 5 ) respectively. Find the resultant of this force system. [05 M]

ans :

\overline{F_1}\;=\;70\;\left[\frac{2i+j+3k}{\sqrt{2^2+1^2+3^2}}\right]

= 37.416i + 18.708j + 56.12k

\overline{F_2}\;=\;80\;\left[\frac{-i+2j}{\sqrt{-1^2+2^2}}\right]=-35.777i+71.554j

\overline{F_3}\;=\;100\;\left[\frac{4i-j+5k}{\sqrt{4^2\pm1^2+5^2}}\right]=61.721i-15.43j+77.152k

Resul\tan t\;\left(\overset\rightharpoonup F\right)=\overset\rightharpoonup{F_1}\;+\overset\rightharpoonup{F_2}\;+\overset\rightharpoonup{F_3}

=37.416i+18.708j+56.12k-35.777i+71.554j

Resultant = 63.36i+74.323j+133.272k

The resultant of the force system=63.36i+74.832j+133.272k

Q. 4 c) A 75 kg person stands on a weighing scale in an elevator 3 seconds after the motion starts from rest, the tension in the hoisting cable was found to be8300N. Find the reading of the scale, in kg during this interval. Also find the velocity of the elevator at the end of this interval. The total mass of the elevator, including mass of the person and the weighing scale, is 750 kg. If the elevator is now moving in the opposite direction, with same magnitude of acceleration, what will be the new reading of the scale ? [08 M]

Ans :

Time (t) = 3 sec

Initial velocity (u) = 0 m/sec

T=8300 Na ; ∑F = ma,

T-W = 750 x a

8300 – 735 a = 750 x a

a = 1.255 m/s²

v = u + at

v = 0 + (1.255 x 3)

For upward motion

N₁– mg = ma

N₁= ma + mg

N₁= m(a+g)

N₁= 75 ( 1.255 + 9.812 )

N₁ = 830.025 N

N₁ = 84.59 kg

For downward motion, mg

N₂– mg = – ma

N₂= mg – ma

N₂= m(g-a)

N₂ = 75 ( 9.812 – 1.255 )

N₂ = 641.775 N

N₂ = 65.407 kg

In the upward motion the reading on the weighing scale is 84.59 kg, Final velocity at the end 3.765 m/s and the reading on the weighing scale is 65.407 kg in the downward direction.

Q. 5 a) The cylinder B, diameter 400 mm and weight 5 kN, is held in position as shown in figure 12 with the help of cable AB. Find the tension in the cable and the reaction developed at contact C.[04 M]

Ans :

\tan\left(d\right)=\frac{200}{350}

d = 29.74°

c + d = 45°

∴ c = 15.26

By Using Lami’s theorem,

\frac5{\sin\left(180-15.26-45\right)}=\frac{F_1}{\sin\left(90+45\right)}=\frac{F_2}{\sin\left(90+15.26\right)}

F₁ = 4.072 kN

F₂ = 5.555 kN

The magnitude of the tension in the cable and the reaction developed at c are 4.072 kN and 5.555 kN.

Q. 5 b) Find the weight WB so as to have its impending motion down the plane. Take weight of block A as 2 kN. The pin connected rod AB is initially is in horizontal position. Refer figure 13.[05 M]

Ans :

Consider Block A

∑ F_x = 0

F_AB + R_A Cos (60) + 0.25 R_A Cos (30) = 0

F_AB + 0.7165 R_A = 0 ⟶ ①

∑ Fy = 0

-2 + R_A Sin (60) – 0.25 R_A Sin (30) = 0

0.741 R_A = 2

R_A = 2.699 kN ⟶ ②

F_AB = -0.7165 R_A

F_AB = -1.934 kN ⟶ ③

Consider Block B

∑ F_x = 0

-F_AB – R_A Cos (45) + 0.25 RB Cos (45) = 0

F_AB = -0.53 R_B

R_B = 3.649 kN

∑ F_y = 0

-W_B + R_B Sin (45) + 0.25 R_BSin (45) = 0

W_B = 3.225 kN

The weight of the block B is 3.22 kN.

Q. 5 c) two springs, each having stiffness of 0.6 N/cm and length 20 cm are connected to a ball B of weight 50 N. The initial tension developed in each spring is 1.6 N. The arrangement is initially horizontal, as shown in figure14. If the ball is allowed to fall from rest, what will be its velocity at D, after it has fallen through a height of 15 cm ? [05 M]

Ans :

Initial tension = 1.6 N

T = kx ; 1.6 = 0.6 x

x_i= 2.667 cm ⟶ I Initial Deformation )

Free length of the spring = I = 20-X_i = 20-2.667 = 17.333 cm

The length of the spring at D = AD = $\sqrt{20^2+15^2}\;=\;25\;cm\$

Deformation at point D = X_f= 25-17.333 = 7.667 cm

Using work energy principle,

∑ Work done = Change in kinetic energy

Gravitational work + Spring work = $\frac12\times\;m\times\;(V_D^2\;\times V_C^2)\$

mgh\;+\;2\left[\frac12k\left(X_i-X_f\right)\right]\frac12\times50\times\left({V^2}_D-0\right)\

=\;\left(50\times9.812\times15\right)+0.6\left(2.667^2-7.667^2\right)=25\;V_D^2\

=\;7359-31.002=25\;V_D^2\

\;V_D^2=293.12\

V_D = 17.12 cm/sec.

The velocity of the ball at point D is 17.12 cm/sec.

Ans :

\tan\left(d\right)=\frac{200}{350}

d = 29.74°

c + d = 45°

∴ c = 15.26

By Using Lami’s theorem,

\frac5{\sin\left(180-15.26-45\right)}=\frac{F_1}{\sin\left(90+45\right)}=\frac{F_2}{\sin\left(90+15.26\right)}

F₁ = 4.072 kN

F₂ = 5.555 kN

The magnitude of the tension in the cable and the reaction developed at c are 4.072 kN and 5.555 kN.

Q. 5 b) Find the weight WB so as to have its impending motion down the plane. Take weight of block A as 2 kN. The pin connected rod AB is initially is in horizontal position. Refer figure 13.[05 M]

Ans :

Consider Block A

∑ F_x = 0

F_AB + R_A Cos (60) + 0.25 R_A Cos (30) = 0

F_AB + 0.7165 R_A = 0 ⟶ ①

∑ Fy = 0

-2 + R_A Sin (60) – 0.25 R_A Sin (30) = 0

0.741 R_A = 2

R_A = 2.699 kN ⟶ ②

F_AB = -0.7165 R_A

F_AB = -1.934 kN ⟶ ③

Consider Block B

∑ F_x = 0

-F_AB – R_A Cos (45) + 0.25 RB Cos (45) = 0

F_AB = -0.53 R_B

R_B = 3.649 kN

∑ F_y = 0

-W_B + R_B Sin (45) + 0.25 R_BSin (45) = 0

W_B = 3.225 kN

The weight of the block B is 3.22 kN.

Ans :

Initial tension = 1.6 N

T = kx ; 1.6 = 0.6 x

x_i= 2.667 cm ⟶ I Initial Deformation )

Free length of the spring = I = 20-X_i = 20-2.667 = 17.333 cm

The length of the spring at D = AD = $\sqrt{20^2+15^2}\;=\;25\;cm\$

Deformation at point D = X_f= 25-17.333 = 7.667 cm

Using work energy principle,

∑ Work done = Change in kinetic energy

Gravitational work + Spring work = $\frac12\times\;m\times\;(V_D^2\;\times V_C^2)\$

mgh\;+\;2\left[\frac12k\left(X_i-X_f\right)\right]\frac12\times50\times\left({V^2}_D-0\right)\

=\;\left(50\times9.812\times15\right)+0.6\left(2.667^2-7.667^2\right)=25\;V_D^2\

=\;7359-31.002=25\;V_D^2\

\;V_D^2=293.12\

V_D = 17.12 cm/sec.

The velocity of the ball at point D is 17.12 cm/sec.

Q. 6 a) For the truss shown in figure 16, find the forces in members DE, BD and CB.[05 M]

Ans :

\sum M_A^F=0\

\left(2\times4\right)+\left(2\times8\right)+\left(2\times12\right)-\left(R_B\times8\right)=0\

R_B =6 kN

Now,

V_A + R_B = 6

V_A = 6-6 = 0 kN

∑ F_x = 0 ; H_A = 0

∑ F_y = 0

V_A -2-2-2+R_B = 0

V_A + R_B = 6 ⟶ ①

In △ ABE,

\tan\left(\alpha\right)\;=\;\frac{EB}{AB}=\frac58

\alpha=\tan^{-1}\left(0.625\right)=32˚

Taking section DD’

∑ F_x = 0 ; $F_{DE}\cos\left(32\right)+F_{BD}\cos\left(32\right)+F_{CB}=0$ ⟶ ②

∑ F_y = 0

-2+F_{DE}\sin\left(32\right)+F_{BD}\sin\left(32\right)=0

$F_{DE}\sin\left(32\right)-F_{BD}\sin\left(32\right)=2$ ⟶ ③

∑ M_D ,

F_D = 0

F_CB x perpendicular distance of F_CB from D = 0

F_CB = 0 kN ⟶ ④

Solving ②, ③ and ④

F_DE = 1.887 kN

F_BD = -1.887 kN

F_CB = 0 kN

The forces in the members DE, BD and CB are 1.887 kN (Compression), 1.887 kN(Tension), and 0 kN respectively.

Q. 6 b) A particle moves in x-y plane with acceleration components a_x= -3 m/s² and a_y = -16 m/s². If its initial velocity is v₀ = 50 m/s directed at 35° to the x-axis, compute the radius of curvature of the path at t=2 sec.[06 M]

Ans :

At t=0

V₀ = 50 m/sec at 35° to the x-axis

V_x= 50 Cos(35°) = 40.96 m/sec

V_y = 50 Sin(35°) = 28.68 m/sec

Given, a_x= -3 m/s² and a_y = -16 m/s²

Integrating, V_x= -3t + C₁ and V_y = -8t²+28.68

at t= 2 sec, V_x= -3(2)+40.96 and V_y = -8(2)²+28.68

V_x= 34.96 m/sec and V_y = -3.32 m/sec

a_x= -3 m/s² and a_y = -32 m/s²

V=\sqrt{V_x^2+V_y^2}

V=\sqrt{34.96^2+\left(-3.32\right)^2}

V = 35.12 m/sec

Radius of curvature at t = 2 sec

R=\frac{V^3}{\left|V_{xay}-V_{yax}\right|}=\frac{35.12^3}{\left|\left(34.96\times-32\right)-\left(3.32\times-3\right)\right|}=38.38\;m

Q. 6 c) A force of magnitude of 20 kN, acts at point A(3,4,5)m and has its line of action passing through B(5,-3,4)m. Calculate the moment of this force about a line passing through points S(2,-5,3)m and T(-3,4,6)m.[05 M]

Ans :

\bar{F_{1}}=20\left [\frac{ \left ( 5-3 \right )i+\left ( -3-4 \right )j+\left ( 4-5 \right )k }{\sqrt{\left ( 5-3 \right )^{2^{}}+\left ( -3-4 \right )^{^{2}}+\left ( 4-5 \right )^{2}}} \right ]

=5.44i – 19.05j -2.72k kN

Overline M_SF1 = Overline SA_xx Overline F₁ .

\begin{bmatrix}i&j&k\\3.2&4.(-5)&5-3\\5.44&-19.05&-2.72\end{bmatrix}=13.62i+13.6j-68.01k\;kNm

\left|M_3F_1\right|=\sqrt{\left(13.62\right)^2+\left(-13.6\right)^2+\left(-68.01\right)^2}=70.68\;kNm

\widehat{ST\;}\;=\frac{\overline{ST}}{\overline{ST}}=\frac{\left(-3-2\right)i+\left(4+5\right)j+\left(6-3\right)k}{\sqrt{\left(-3-2\right)^2+\left(4+5\right)^2+\left(6-3\right)^2}}=0.466i+0.839j+0.28k

Moment about the line,

M_{ST}F_1=M_sF_1.\widehat{ST}\;=\;\left(13.62i+13.6j-68.01k\right).\left(-.0466i+0.839j+0.28k\right)

=\;-6.35+11.41-19.04

M_{ST}F_1\;=-13.98\;kNm

Vector form

M_{ST}F_1=M_sF_1.\widehat{ST}\;=-13.98\left(-0.466i+0.839j+0.28k\right)\;kNm

M_{ST}F_1=6.51i-11.73j-3.9k

The moment of the force about line passing through points S(2,-5,3)m and T(-3,4,6)m is -13.98 kNm magnitude and 6.51i-11.73j-3.91k

Q. 6 d) Find an expression for maximum range of a particle which is projected with an initial velocity of ‘u’ inclined at an angle of ‘β’ with the horizontal.[04 M]

Ans :

Consider a pentacle performing projectile motion

R → Horizontal range ; T → Total flight time

Considering vetical components of motion

S=ut+\frac12at^2\;

Ou\sin\left(\beta\right)-\frac129T^2

T=\frac{2u\sin\left(\beta\right)}9

Considering horizontal components of motion

S=ut+\frac12at^2\;

R\;=\;u\cos\left(\beta\right)T+0\;\rightarrow\;\left(acc.\;in\;x\;direction\;is\;0\right)

R\;=\;u\cos\left(\beta\right)\times\frac{2u\sin\left(\beta\right)}9

R\;=\;\frac{u^2\sin\left(2\beta\right)}9

For maximum range, sin(2β) should be max. i.e. sin(2β)=1

i.e. 2β = 90°

β = 45°

R_{max}\;=\;\frac{u^2}9