﻿ Statics forces solved problems
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 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 Then if the weight is given it equals W [kgf ] = 9.8 N
 Static forces example - 1 ▲   m kgm M kgm α deg θ deg T N Fx N Fy 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: Fx − T sinθ = 0 (1) y direction: Fy + 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 ΣMA = 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 Fx and Fy : (5) (6)
Once we found  Fx  and  Fy  we can find the reaction force  F  and the angle of the reaction  γ  on hinge  A.  If  α + θ = 90 degree  then  T  Fx  and  Fy  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 Fx is: Fx = sin90 sin60 (500 + 10 / 2) / sin150 = 874.7 kgf Fy is: Fy = 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 Fx N Fy 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 Fx − T cosθ = 0 (1) y direction Fy + 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 Fx Fy 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: 