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Concurrent
forces, weight, tension, friction |
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Equilibrium
Conditions |
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Problem Solution |
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Kinetic friction, Static friction, coefficients and laws of friction |
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Information
- Define the following terms?
| Concurrent forces: Are forces whose lines of action all pass through a
common point. |
| Weight: is the force with which the gravity pulls downward upon it. |
| Tension in string: is the force with which the strings pulls upon the
object to which it is attached. |
| Friction force: is a tangential force on a surface that opposes the
sliding of a surface across an adjacent surface. |
| Normal force: is the perpendicular component of the force exerted by
the supporting surface on the surface being supported. |
- What are the equilibrium conditions under the action of concurrent forces?
The resultant of all forces acting on an object
must be zero. or
1- The sum of all x-components is zero.
2- The sum of all y-components is zero.
3- The sum of all z-components is zero.
- When an object is in equilibrium
If it is at rest and remains at rest. or if it
is in motion with constant vector velocity
- What are the types of equilibrium
Static-Equilibrium: The object it is at rest
and remains at rest.
Translational-Equilibrium: The object is in motion with constant vector
velocity
- How do you solve problems in equilibrium under
concurrent forces?
1- Isolate the object under forces
2- Draw a free-body diagram
3- Find all the x,y,z-components of each force in the system
4- Write the conditions for equilibrium in an equation form
5- Solve for the unknowns from the equations obtained in step-4.
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Problems
- The system shown below is in equilibrium. Solve for T1 and T2.
1- SFx=0,
-T1Sin30+T2Sin60=0, -0.5T1+0.866T2=0
T1=1.73T2
2- SFy=0, T1Cos30+T2Cos60-20=0,
0.866T1+0.5T2=20
From (1) 0.866(1.73T2)+0.5T2=20, 1.50T2+0.5T2=20
T2=20/2=10 N
T1=1.73T2=17.3 N |
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- Two forces, P and Q, act NW and NE respectively. They are in
equilibrium with a force of 50.0 N acting due E. Find P and Q.
1- SFx=0,
QSin45+20-PSin45=0
0.7Q-0.7P=-20
2- SFy=0,
QCos45+PCos45-50=0
0.7Q+0.7P=50
By adding (1) & (2) 1.4Q=30, Q=30/1.4=21.4 N
From (1) P=(Q+20/0.7)=49.9N |
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- A particle whose weight is 50N is suspended by alight string which is at
35 degrees to the vertical under the action of a horizontal force F.
Find: (a) the tension in the string, (b) F.
1-
SFx=0, F-TSin35=0
2- SFy=0, TCos35-50=0
From (2) T=50/Cos35=61N
From (1) F=TSin35=61Sin35=35 N |
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- A particle of weight W rests on a smooth plane which is inclined at 40
degrees to the horizontal. The particle is prevented from slipping by
a force of 50N acting parallel to the plane and up a line of greatest
slope. Calculate (a) W, (b) the reaction due to the plane.
1- SFx=0,
50-wSin40=0
W=50/Sin40=77.8 N
2- SFy=0, R-wCos40=0
R=wCos40=59.6 N
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- Two light strings are perpendicular to each other and support a particle
of weight 100N. The tension in one of the strings is 40N.
Calculate the angle this string makes with the vertical and the tension in
the other string.
1- SFx=0,
TCosa-40Cosq=0
2- SFy=0, 40Sinq+TSina-100=0
Since q+a=90,
we have
Sinq=Cosa
and
Cosq=Sina
1- TSinq-40Cosq=0
2- TCosq+40Sinq=100
Multiply (1) by
Cosq
and (2) by
Sinq.
1- -TCosqSinq+40CosqCosq=0
2- TSinqCosq+40SinqSinq=100Sinq
40(Sin2q+Cos2q)=100Sinq,
Sinq=0.4, q=23.5,
a=90-q=90-23.5=66.4
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Notes:
1) When q+a=90, then
Sinq=Cosa
Cosq=Sina
2)Sin2q+Cos2q=1
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