Statics forces solved problems

Notes, about the differences between mass and weight

According to the second law of motion we have	F = m a
In the earth gravitational system the force acting on a mass is	W = m g [kgm m / s² ] = 1 N
A force of 1 Newton is equal to	N = 1 [kgm m / s² ] = 0.102 kgf

The force of 1 kgf is equal to:	1 kgf = 9.8 N

The mass is a fundamental property of the body which changes in different gravitational environments thus the weight W of a body equals to the mass m multiply by the gravity g.	W = m • g

Static forces example - 1

m		kgm
M		kgm
α		deg
θ		deg
T		N
F_x		N
F_y		N
F		N
γ		deg

A beam AB whose weigh is M and is located at an angle of α from the wall and is hinged at point A to the wall, a cable is connected to the wall and the beam end at point B as shown in the figure. Find the direction and the value of the force acting on hinge A and the tension in the cable CB.

From the free body diagram, we can write the equilibrium equations in the x and y directions:

x direction:	F_x − T sinθ = 0	(1)
y direction:	F_y + T cosθ − mg − M g = 0	(2)

Another condition that should be fulfilled is that the sum of the moments at the hinge A is ΣM_A = 0

If α + θ is not 90 degree then we have to calculate the projection of T that is perpendicular to the beam which is T sin(α + θ)

Moments at A:

m g L sinα + M g sinα L / 2 − T sin(α + θ) L = 0

(3)

from eq. (3) we can find the value of T.

	(4)
Substitute eq. (4) into eq. (1) and (2) we get the forces F_x and F_y :
	(5)
	(6)

Once we found F_x and F_y we can find the reaction force F and the angle of the reaction γ on hinge A.

If α + θ = 90 degree then T F_x and F_y are:

The weight of a uniform horizontal beam is 10kgf a weight of
m = 500kgf is hanged on point B . A cable holds the end of the beam to the wall at an angle of 60 degree. Find the reaction force on hinge A and its direction. (run example).

Because α + θ = 150 degree we have to use eq. (4) (5) and (6).

Given the weight W which is equal to m g = W

Tension is:	T = sin 90 (500 + 10 / 2)/cos(90 + 60 − 90) = 1010 kgf
F_x is:	F_x = sin90 sin60 (500 + 10 / 2) / sin150 = 874.7 kgf
F_y is:	F_y = 510 − sin90 cos60 (500 + 10 / 2) / sin150 = 5 kgf
γ is:	γ = arctan (874.7 / 5) = 89.7 degree

Static forces example - 2

m		kgm
α		deg
β		deg
T₁		N
T₂		N

From the free body diagram sum of all forces in the x and y directions should be equal to 0.

x direction	T₁ cosα − T₂ cosβ = 0
y direction	m g − T₁ sinα − T₂ sinβ = 0

Solving for T₁ and T₂ we get:

Static forces example - 3

m		kgm
θ		deg
α		deg
T₁		N
T₂		N

Given the value of the mass and the angles θ and α as shown in the figure. Find the tension in the cables holding the mass.

Sum of all forces in the x and y directions should be equal to 0.

In the x direction:	T₂ cosθ − T₁ cosα = 0	(1)
In the y direction:	m g + T₁ sinα − T₂ sinθ = 0	(2)

From eq. (1) and (2) we can find the values of T₁ and T₂

Note: angle θ should be bigger then angle α (θ > α) otherwise T₁ is

negative, but the cable can hold only positive tension.

When α = 0 (cable is horizontal) the equations of T₁ and T₂ are.

Static forces example - 4

W		kgm
M		kgm
θ		deg
a		m
b		m
T		N
F		N
F_x		N
F_y		N
γ		deg

Beam AB is supported by a cable to the wall and makes an angle of θ with the beam, a weight W [kgf] is suspended from the beam at a distance a from the wall as shown in the figure. Determine the tension T in the supporting cable and the direction and value of the force on the pin at point A.

From equilibrium all forces and moments in x and y directions are 0.

x direction	F_x − T cosθ = 0	(1)
y direction	F_y + T sin θ − W − M = 0	(2)
Moments at A	a W + M (a + b) / 2 − (a + b) T sinθ = 0	(3)

We got three equations with F_x F_y and T as unknowns, solving by matrix form we get.

From eq. (3) we can find the tension in the cable.

And the force angle γ is:

Static forces example - 5

Two masses m₁ = 3kg and mass m₂ = 2kg are hanging on a rope in equilibrium as shown in the figure. Find a) the angle γ if we know the values of the masses and the angles α = 45° and β = 45°. b) what is the ratio of m₂ / m₁ which results the angle γ to be γ = 53.435° c) Find the tension in cable 3 as a function of m₁ and g if the ratio m₂ / m₁ is the value found in section b.

From the forces diagram we see that:

T₄ = m₁ g

T₅ = m₂ g

Forces diagram of m₁

Forces diagram of m₂

From forces balance in the x and y directions of mass m₁

∑F_x:

T₁ cosα = T₂ cosβ

(1)

∑F_y:

m₁ g = T₁ sinα + T₂ sinβ

(2)

From forces balance in the x and y directions of mass m₂

∑F_x:

T₂ cosβ = T₃ cosγ

(3)

∑F_y:

m₂ g + T₂ sinβ = T₃ sinγ

(4)

Equations 1) and 2) can be written in linear equations matrix form and solved by Cramer’s law.

(■(cos⁡α&-cos⁡β@sin⁡α&sin⁡β))(■(T_1@T_2))=(■(0@m_1 g)) Friction example 5

Solving equations 1 and 2 for the unknowns T₁ and T₂

T_1=T_2 cos⁡β/cos⁡α =(m_1 g cos⁡β)/(2 sin⁡α cos⁡α) Friction example 5

Divide equation 4 by equation 3 to get:

tan⁡γ=(m_2 g+T_2 sin⁡β)/(T_2 cos⁡β ) Friction example 5

Substitute T₂ from the previous equation we get:

tan⁡γ=1+m_2/m_1 (sin⁡β+tan⁡α cos⁡β ) Friction example 5

(5)

From the trigonometric angle values we get from the question that tanα = tanβ = 1 and the γ angle is:

b) From equation (5) we have:

m_2/m_1 =(tan⁡γ-tan⁡β)/(tan⁡β+tan⁡α )=tan⁡〖63.43-tan⁡45 〗/(tan⁡45+tan⁡45 )=0.5

c) From equation (3) we have:

Entering T₂ into equation (4):

From the ratio of m₂/m₁ = 0.5 we get: m₂ = m_/2

And T₃ is:

T_3=(m_1 g)/2(sin⁡(63.43)-cos(⁡63.43)/cos⁡45)=1.12m_1 g

Static forces example - 6

Forces diagram

A uniform bar of length l is placed on a tube at a distance of x from the edge of the bar which is leaning on the wall at an angle of θ so that the bar is at rest. Draw all the forces acting on the bar and find the distance x that will cause the bar to be at equilibrium if the length of the bar is 0.8 m and the weight of the bar is 10kgf and the angle θ = 60°.

There are three forces acting on the bar:

1. The weight of the bar W acting in the centre of gravity.

2. The reaction of the force where the bar is resting on the circular tube.

3. The reaction of the force of contact of the bar and the wall.

The bar is at rest so all those forces are at equilibrium in x and y directions:

∑F_x = 0:

N − N₁ cosθ = 0

(1)

∑F_y = 0:

W − N₁ sinθ = 0

(2)

∑M_A = 0:

N₁ x - W sinθ L / 2 = 0

(3)

From equations (2) and (3) we can find the value of x

x=l/2 sin^2⁡θ=0.8/2 sin^2⁡60=0.3 m

From equation (3) we can find N₁:

N_1=WL/2x sin⁡θ=(10*0.8)/(2*0.3) sin⁡60=11.54 kgf

Divide equation (2) by (1) we get N:

N = W / tanθ = 10 / tan60 = 5.77 kgf

Forces diagram including friction

If there is friction between the bar and the tube and this time the value of x is changed to 0.2m the angle θ is the same 60° and the bar is at rest, find the friction force.

The total forces acting in the x and y directions and the moment around point A are:

∑F_x = 0:

N − N₁ cosθ − f sinθ = 0

(1)

∑F_y = 0:

W − N₁ sinθ − f cosθ = 0

(2)

∑M_A = 0:

N₁ a - W sinθ L / 2 = 0

(3)

From equation (3) we can find N₁:

N_1=(W L)/(2 a) sin⁡θ=(10*0.8)/(2*0.2) sin⁡60=17.32 Kgf

Friction force is found from equation (2):

f=(W-N_1 sin⁡θ)/cos⁡θ =(10-17.32 sin⁡60)/cos⁡60 =-10 kgf

The negative value of the friction is indicating that the direction of the force is in the opposite direction from the direction selected.

Static forces example - 7

Free body diagram W/O friction

A bar is tied with a rope to the wall at an angle of θ degree. Draw all the forces acting on the bar and find the tension in the rope as a function of angle θ and the weight of the bar W. Assume that there is friction between the bar and the floor. And find the minimum friction force needed to keep the system in balance.

Forces balance in the x and y directions are:

∑F_x = 0:

T − f = 0

(1)

∑F_y = 0:

W − N = 0

(2)

The moments balance around the center of gravity is:

∑M = 0:

T sinθ L/2 + f sinθ L/2 − N cosθ L/2 = 0

T sinθ + f sinθ − N cosθ = 0

(3)

Substitute equations (1) and (2) into equation (3) to get:

f=(W cos⁡θ)/(2 sin⁡θ)=W/(2 tan⁡θ)

The friction force is

f = N μ

The friction coefficient should be bigger then

μ≥f/N=W/(2 tan⁡θ W)=1/(2 tan⁡θ)

Static forces example - 8

Free body diagram

W		kgm
θ		deg
μ_S
F		N
T		N
N		N

Notice: Friction coefficient should be less then
μ_S < tanθ

Vertical bar of a weight of W is placed on the floor with friction of μ and is tied to the floor with a rope at an angle of θ. Draw the forces acting on the bar, what is the maximum force F that the system will stay at rest and find the tension in the rope at that state.

Forces balance in the x and y directions are:

∑F_x = 0:

T sinθ + f − F = 0

(1)

∑F_y = 0:

W + T cosθ − N = 0

(2)

The moments balance around the center of gravity is:

∑M_B = 0:

F L / 2 − T sinθ L = 0

(3)

The friction force between the floor and the vertical bar is:

f = N μ_S

(4)

The value of f from eq. (4) insert into eq. (1) to get:

∑F_x = 0:

T sinθ + N μ_S − F = 0

(5)

We can write equations (2), (3) and (5) in matrix notation, with unknowns T, N and F and solve by Cramer’s rule.

(■(cos⁡θ&-1&0@L sin⁡θ&0&-L/2@sin⁡θ&μ_S&-1))(■(T@N@F))=(■(-W@0@0))

The determinant of the coefficients is:

T=|■(-W&-1&0@0&0&-L/2@0&μ_S&-1)|/D=(Wμ_S)/(μ_S cos⁡θ-sin⁡θ )

N=|■(cos⁡θ&-W&0@L sin⁡θ&0&-L/2@sin⁡θ&0&-1)|/D=(W sin⁡θ)/(2 sin⁡θ-μ_S cos⁡θ )

F=|■(cos⁡θ&-1&-W@L sin⁡θ&0&0@sin⁡θ&μ_S&0)|/D=(2Wμ_S sin⁡θ)/(〖2sin⁡θ-μ〗_S cos⁡θ )

Static forces example - 9

W		kgf
W_L		kgf
L		m
a		m
θ		deg
T₁		N
F_x		N
F_y		N

Notice: Friction coefficient should be less then
μ_S < tanθ

A weight of 380 kgf is hanging on a crane whose weight is 200 kgf and a length of 100 m at an angle of 60 degree. Find the stress on cable A-C if the distance a is 40 m.

Write the sum of the moments around point B.

∑M_B = 0:

T₁ h − W L cosθ − W_L cosθ L / 2 = 0

(1)

Now the only problem is to find the value of h this we can find from trigonometry by cosine law. First, we will find the value of b:

b² = L² + a² − 2 L a cos(180 − θ)

The value of cos(180 − θ) = −cosθ

Now we can find the value of angle β by sin law

b/sin⁡(180-θ) =a/sin⁡β

β=sin^(-1)⁡((a sin⁡(180-θ))/b)

And h is equal to: h = L sinβ

The reaction forces on the hinge connecting the bar to the floor can be found by forces equilibrium.

∑F_x = 0:

F_x − T₁ sin(β + γ) = 0

(2)

∑F_y = 0:

F_y − m g − W − T₁ cos(β + γ) = 0

(3)

The angle γ is equal to 90 − θ

Find the value of cable length b

b=√(100^2+40^2+2∙100∙40 cos⁡60 )=124.9 m

β=sin^(-1)⁡((40 sin⁡60)/124.9)=16.1°

h=100 sin⁡〖16.1=27.735 m〗

T_1=(100∙cos⁡60 (2∙380+200))/(2∙27.735)=865.3 kgf

Static forces example - 10

Two masses M of 40 kg each are hanging by two ideal pullies who has no friction. Another mass m of 50 kg is hanging on the middle of the cable. Draw the forces acting on the system and find the angle α.

Forces acting on both masses are:

∑F_y (M) = 0:

T − M g = 0

(1)

∑F_y (m) = 0:

m g − 2 T sinα = 0

(2)

From equation (1) we can find the value of T and from equation (2) we can find the value of the angle α.

sin⁡α=mg/2T=mg/rMg=m/2M

α=sin^(-1)⁡〖m/2M=sin^(-1)⁡〖50/(2∙40)〗〗=sin^(-1)⁡0.625=38.7°

Notice that the relation m < 2M must be true otherwise the system will not be in equilibrium.

M		kgf
m		kgf
α		deg