Concept:

Component of ${\overrightarrow{F}}_2$ in ${\overrightarrow{F}}_1={\overrightarrow{F}}_2\cos \theta$

$\mathrm{W}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e},\cos \theta =\frac{{\overrightarrow{F}}_1.{\overrightarrow{F}}_2}{\left|{\overrightarrow{F}}_1\right|\left|{\overrightarrow{F}}_2\right|}$

Calculation:

${\overrightarrow{F}}_1.{\overrightarrow{F}}_2=4+6+\left(-3\right)=7$

$\left|{\overrightarrow{F}}_1\right|=3.74N;\left|{\overrightarrow{F}}_2\right|=5.099N$

cos θ = 0.367

∴ F₂ cos θ = 5.099 × 0.367

∴ F₂ cos θ = 1.87 N

Explanation:

When we draw free body diagram we remove all the supports and force applied due to those support are drawn and force or moment will applied in that manner so that it resist force in any direction as well as any tendency of rotation.

Following is the conclusion:

Option (B) is wrong because ground forces are shown which we never show in FBD.

Option (C) is wrong because X component is not shown.

Option (D) is wrong because moment produced due to the eccentric force is not shown.

Given: T = 20 N, μ = 0.25, cos θ = 0.8, sin θ = 0.6, g = 10 m/s²

Friction always acts opposite to the direction of motion.

N = mg cos θ = m × 10 × 0.8 = 8 m

∵ It is given static friction. Therefore there will be no motion i.e. a = 0

mg sin θ = m × 10 × 0.6 = 6 m

mg sin θ = μ_sN + T

6 m = 0.25 × 8 m + 20

Concept: Using Newton’s second law.

Calculation:

Free body diagram of above mass-pulley system

$m_{1}g=100N\Rightarrow m_1=\frac{100}{g}=10.19kg$

$m_{2}g=200N\Rightarrow m_2=\frac{200}{g}=20.38kg$

Consider block of 100 N:

From Newton’s 2^nd law

T - 100 = m₁a

∴ T = 10.19 a + 100 …i)

Now, Consider block of 200 N:

∴ 200 - T = m₂a

∴ T = 200 - 20.38 a      …ii)

From equation i) & ii)

10.19 a + 100 = 200 - 20.38 a

⇒ a = 3.27 m/s²

Now, substituting the value of a in equation (ii)

T = 200 - 20.38 × (3.27)

Concept:

FBD of Body A

B = 1500 N

A = 1000 N

∑F_y = ma

$1000-T=\frac{1000}{10}a$     ---(1)

FBD of Body B,

∑F_x = ma

$T-\mu N=\frac{1500}{10}a$      ---(2)

∑F_y = 0

∴ N = W = 1500 N

Now,
Using equation (1) and (2)

1000 – T = 100 a

And T – (0.2 × 1500) = 150 a

Adding,

1000 – 300 = 250 a

$\therefore \frac{700}{250}=a$

∴ a = 2.8 m/s²

Explanation:

∑M_B = 0

R_A × 4 – 5 × 3 – 10 × 1 = 0

∴ R_A = 6.25 kN

Using method of section,

∑M_F = 0

F_CD × 2 sin 60 – 5 × 1 + 6.25 × 2 = 0

$\therefore F_{CD}=\frac{5\times 1-6.25\times 2}{2\sin {60}^{\circ }}$

F_CD = -4.33 kN

∴ F_CD = 4.33 kN (Compression)   

Concept:

• When cylinder is on the verge of rolling over the block, the contact with the ground is lost and hence normal reaction (from the ground) will not develop.
• If only 3 coplanar forces are acting on a body to keep it in equilibrium, then they must be concurrent.

Calculation:

Case: (a)

Given:

W = 5000 N

Now,

Applying moment about O

P × 400 = W × 447.21 (1)

P = 5590.16 N (option a)

Case: (b)

Three forces have to be concurrent, so N will pass through the top point.

Applying moment about O

P × 1000 = W × 447.21

∴ P = 2236.07 N (option b)

Concept:

Free Diagram of Body (FBD) is made to solve easily

Calculation:

Given,

Weight of Block A (W_A) = 800 N, Weight of block B (W_B ) = 600 N

μ _A = 0.25, μ _B = 0.33

i) FBD of Block (W)

From FBD, W = T

ii) FBD of the body

From FBD

$T=T_A\times \sin {45}^{\circ }+T_B\times \sin {60}^{\circ }$

∴ Weight of Block

$W=T_A\times \sin {45}^{\circ }+T_B\times \sin {60}^{\circ }$

iii) FBD of Block A

From FBD,

$T_A+f_a=W_A\sin \alpha$

Where f_a is frictional force = μ_A × N_A

$N_A=W_A\times \cos {25}^{\circ }$

$\begin{array}{l}800\sin {25}^{\circ }=0.25\times 800\cos {25}^{\circ }+T_A\\ T_A=800\sin {25}^{\circ }-0.25\times 800\cos {25}^{\circ }\end{array}$

= 156.833 N

FBD of block B

From FBD

$T_B+f_b=W_B\times \sin \alpha$

Where f_b is frictional force = μ_B × N_B 

$\begin{array}{l}{N}_B=W_B\times \cos {33}^{\circ }\\ T_B=600\times \sin {33}^{\circ }-0.33\times 600\times \cos {33}^{\circ }\end{array}$

= 160.726 N

∴ Weight of block

$W=T_A\times \sin {45}^{\circ }+T_B\times \sin {60}^{\circ }$

$W=156.83\frac{1}{\sqrt{2}}+160.726\times \frac{\sqrt{3}}{2}$

W = 250.090 N.

Concept:

Free body diagram of Roller (1) and (2)

From FBD of Roller (1)

$\frac{W}{\sin {90}^{\circ }}=\frac{N_1}{\sin {120}^{\circ }}=\frac{N_3}{\sin {150}^{\circ }}$

$\frac{450}{\sin {90}^{\circ }}=\frac{N_1}{\sin {120}^{\circ }}=\frac{N_3}{\sin {150}^{\circ }}$

∴ N₁ = 389.7 N

N₃ = 225 N

From FBD of Roller (2)

∑F_xx = 0

R_A cos 30 – N₃ – W cos 60 = 0

$R_A=\frac{N_3+W\cos 60}{\cos 30}$

∴ R_A = 519.6 N 

Concept:

If in truss problem there are 3 members in which 2 are in the same line and there is no external force, then the force in the 3^rd member will be zero

Calculation:

Consider point B;

F_BC = 0

Consider point C;

∵ F_BC is already zero

F_AC and F_CE is in the same line so, F_DC = 0