We Have Numbers Of Free Samples

Solution 1:

a.

Given data:
Couple moment, M: 4 Nm

To find:
F and  P

Now,

C.M = F.d

Where,
F is the Force applied
M is the Moment
d is the distance

4=F (0.03)
F=133 N

Similarly:

The force F is found to be 133 N and the Force P is calculated to be 800 N

b.

Given:

F: 5 KN
F = 6 KN

To find:

Reaction forces at fixed joint A.
A_x and A_y

Now,
Summation of Horizontal forces:

Summation of Vertical forces:

A_y = 0 KN

Summation of Moments:

For F = 5 KN

For, F = 6 KN

M₂ = -3.8 KN m

The Resultant can be found as:

M_R = M1 + M2
M_R – 5.2 KN.m

Solution 2:

Now,

To find:

a. Reaction forces at the support

• R_Ay
• R_By
• R_Bx

b. Magnitude of forces acting in member BD, and ED.

a.

Now considering the summation of Horizontal forces:

Considering the summation of Vertical Forces:

Taking Moment About A:

2 (800) + (2 + 2 cos⁡60 ) (200) – R_Ey = 0
R_Ey = 2200 N

Now,

R_Ay + R_Ey = 1000 N
R_Ay = -1200 N

b. Magnitude of Forces acting in member BD, and ED.

At BD:

Summation of Horizontal forces

F_BD = 150 + ve Tensile

Summation of Vertical forces

Fy = -1200 + 2200 + 800 + 200 = 2000 N (Tensile)

At, ED:

Summation of Horizontal forces

F_BD = 150 + ve Tensile

Fy = -1200 + 2200 + 800 + 200 = 2000 N (Tensile)

Solution 3:

Support Reaction calculations:

ΣFx = 0

HA = 0

The sum of the moments about the pin support at the point A:

ΣMA = 0

The sum of the moments about the roller support at the point B:

ΣMB = 0

– RA*6 + q1*6*(6 – 6/2) – M1 – P1*2 = 0

Calculation of reaction of roller support at the point B:

RB = ( q1*6*(6/2) + M1 + P1*8) / 6 = ( 4*6*(6/2) + 6 + 3*8) / 6
RB= 17.00 (kN)

Calculation of reaction of pin support at the point A:

RA = ( q1*6*(6 – 6/2) – M1 – P1*2) / 6
RA = ( 4)6(6 – 6/2) – 6 – 3(2) / 6 = 10 (kN)

Consider first span of the beam:

Span:  0 ≤ x1 < 3

Determine the equations for the axial force (N):

N(x1) = HA

The values of N at the edges of the span:

N1(0) = 0 (kN)
N1(3) = 0 (kN)

Determine the equations for the shear force (Q):

Q(x1) = + RA – q1*(x1 – 0)

The values of Q at the edges of the span:

Q1(0) = + 10 – 4*(0 – 0) = 10 (kN)
Q1(3) = + 10 – 4*(3 – 0) = -2 (kN)

The value of Q on this span that crosses the horizontal axis. Intersection point:

x = 2.50

Determine the equations for the bending moment (M):

?(?1) = + ??∗(?1) − ?1∗(?1)2/2

The values of M at the edges of the span:

M1(0) = + 10*(0) – 4*(0 – 0)2/2 = 0 (kN*m)
M1(3) = + 10*(3) – 4*(3 – 0)2/2 = 12 (kN*m)

Local extremum at the point x = 2.50:

M1(2.50) = + 10*(2.50) – 4*(2.50 – 0)2/2 = 12.50 (kN*m)

Considering second span of the beam

Span: 3 ≤ x2 < 6

Determine the equations for the shear force (Q):

Q(x2) = + RA – q1*(x2 – 0)

The values of Q at the edges of the span:

Q2(3) = + 10 – 4*(3 – 0) = -2 (kN)
Q2(6) = + 10 – 4*(6 – 0) = -14 (kN)

The values of M at the edges of the span:

M2(3) = + 10*(3) – 4*(3 – 0)2/2 + 6 = 18 (kN*m)
M2(6) = + 10*(6) – 4*(6 – 0)2/2 + 6 = -6 (kN*m)

Considering the third span of the beam

Span: 6 ≤ x3 < 8

The values of Q at the edges of the span:

Q3(6) = + 10 – 4*(6 – 0) + 17 = 3 (kN)
Q3(8) = + 10 – 4*(6 – 0) + 17 = 3 (kN)

The equations for the bending moment (M):

M(x3) = + RA*(x3) – q1*(6 – 0)*[(x3 – 6) + (6 – 0)/2] + M1 + RB*(x3 – 6)

The values of M at the edges of the span:

M3(6) = + 10*(6) – 4*6*(0 + 3) + 6 + 17*(6 – 6) = -6 (kN*m)
M3(8) = + 10*(8) – 4*6*(2 + 3) + 6 + 17*(8 – 6) = 0 (kN*m)

Solution 4:

Given:

Mass,

Ma = 2500 N

Mb = 4000 N

cof = 0.28

K.cof = 0.25

a) Tension force in Rope connected to Crate A

tan⁡θ=3/4
θ=36.86

F=sin 36.86 (2500 )=1469 N

b) Tension force in B

tan⁡θ=3/4
θ=36.86
F=sin 36.86 (4000 )=2400 N

C)

P required to move Crate B at 0.8m/s2

P=Ma

The Required force to move the crate B is 4070 N

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