6.1 Equilibrium of Particles
> Any object is said to be in equilibrium if its acceleration is zero.
> If the velocity is also zero, the object is said to be in static equilibrium.
> Since F = ma, if a = 0, then F = 0. Thus the conditions for a point object (particle) in equilibrium is:
the vector sum of all the external forces acting on the particle must be zero, ΣF = 0
> For forces in a plane, (x-y plane) in order that a particle is in equilibrium, the sums of the resolved parts of the forces in the x and y directions must be zero. ΣF_{x} = 0 and ΣF_{y} = 0. (In general, to prove that a particle is in equilibrium, we must show that the sums of the resolved parts of the forces on the particle in any two directions are each equal zero.)
6.2 Closed Polygon
> If the forces as mention in 6.1, drawn to scale and in the appropriate directions, a close triangle or a closed polygon is formed.
For example,(summary)
Example 1
Coplanar forces F_{1},F_{2} and F_{3} act on a point mass A. as shown in figure above. State or show on a labeIled diagram, the condition for the mass to be in equilibrium.
solution
Example 2
a small ball with weight W = 20 N is suspended by a light thread. When strong wind blows horizontally, exerting a instant force F on the ball, the thread makes an angle 30° to the vertical as shown. What are the values of T and F ?
solution
Example 3
A cable car travels along a fixed support cable and is pulled along this cable by a moving draw cable. For the situation shown, the cable car and passengers with weight 50 x 10^{4} N is considered to be stationary with forces T, and T, acting on it. Assuming the draw cable exerts negligible force on the cable car.
(a) sketch and show the forces acting on the cable car.
(b) find the magnitude of T_{1} and T_{2}
Solution
Example 4
An object of weight W is placed on a smooth plane inclined at an angle q to the horizontal. A force F is applied in a direction parallel to the plane so that the body is in equilibrium. Find F and R, the I normal reaction in terms of W and q. ,
Solution
Resolving forces in a direction perpendicular to the plane, R = W cos q
Resolving forces in a direction parallel to the plane, F = W sin q
Example 5
An object of weight W is placed on a smooth plane inclined at an angle 0 to the horizontal. A force F is applied horizontally so that / the body is in equilibrium. Find F and R, the normal reaction in / terms of W and 0.
Solution
Turning effects of forces
• Torque or moment of a force about a point is the product of that force and the perpendicular distance from the line of action of the force to the point.
(a)Torque/Moment, t = F x d
(b) Torque/Moment = F d sin q
• moment of a force = force x perpendicular distance from pivot to the line of action of force
• Unit of moment is Newton metre (N m)
· torque is a vector quantity
• A torque can turn in a clockwise direction or an anticlockwise direction.
Example 6
The crank of a bicycle pedal is 16 cm long and the downwards push of a leg is 300 N. Calculate the moment due to the force exerted on the pedal when crack has turned through an angle of 60° below the horizontal.
Solution
Moment of the force = 300 x 0.16 cos 60° = 300 x0.16 x0.5 = 24 Nm
Couple/Gandingan
•When two equal forces are acting in opposite direction but not along the same straight line, they form a couple.
•A couple has no resultant force.
•A couple only produces turning effect (rotation).
•The moment of a couple is given by:
• The moment of a couple is the product of one of the forces and the perpendicular distance
between the two forces.
Summary
Torque/Moment* | Couple |
Turning Effect
t = F x d F : force d : jejari/radius Moment = F d sin q Equilibrium when moment clockwise = anticlockwise
*in physic moment of force(moment) and Torque, they have the same meaning |
When two equal forces are acting in opposite direction |
Example 7
Calculate the moment of the Couple produced by the forces acting on the steering wheel of a car in the diagram given.
Solution
Moment of couple = 25 x 0.30 = 7.5 N m
6.3 Equilibrium of Rigid Bodies
> For a rigid body (a non-point object or extended object) in equilibrium,
(a) there must be zero resultant force. ΣF = 0
(b) there must be zero resultant torque. Στ = 0
Alternatively, we can apply the principle of moments which state that for any body in equilibrium, the sum of the clockwise moments about any pivot must equal the sum of , the anticlockwise moments about that pivot.
> Forces that act on rigid body may be concurrent forces or non-concurrent forces.
1.concurrent forces(Daya bersetemu)
– Concurrent forces are forces whose lines of action pass through a single common point.
– Concurrent forces will only cause translational motion.
– To determine whether concurrent forces acting on a rigid body are in equilibrium, all that is required is to check whether the resultant forces is zero.
2. Under non concurrent forces
– Non-concurrent forces are forces whose lines of action do not pass through a single common point.
– To determine whether non-concurrent forces acting on a rigid body has translational and rotational equilibrium, it is necessary to check whether condition ΣF = 0 as well as condition Στ = 0.
Example 8
The seesaw in the diagram is balanced. Use the principle of moments to calculate the weight, W.
Solution
W(15) = 300(1.0) + 450(1.5) = 650N
Example 9
uniform rod XY of weight 20.0 N is freely hinged to a wall at X. It is held horizontal by force F acting from Y at an angle of 60° to the vertical as shown in the diagram. What is the value of F?
solution
The diagram shows the forces acting on your forearm when you hold a weight of 60 N with your arm horizontally. Your elbow joint acts as a fulcrum. The weight of the arm is 20 N and its
is considered to act at a distance 14.0 cm from the fulcrum. Use the principle of moments to calculate the force T exerted by your biseps. What is the reaction force R at the fulcrum?
solution
6.4 Frictional Forces
> Friction acts whenever two surfaces move or try to move relative to one another.
> There are two kinds of frictional forces:
1. Static friction
– It is a force at the contact surfaces which prevents the surfaces from sliding over each other.
-The frictional force always acts in the opposite direction to the pulling force P. It is always self-adjusting, constantly equalising itself to P, maintaining static equilibrium as long as the limiting friction is not exceeded.
-If the pulling force is greater than the limiting friction, the block moves and anoth frictional force known as kinetic friction comes into effect.
Limiting friction:
(a) Depands on the nature of the surfaces.
(b) Is independent of the area of contact.
(c) Is proportional to the normal reaction,R.
Thus limiting friction has a value F = mR
2.Kinetic friction
-It is a force between two moving surfaces which opposes the sliding motion.
– When the pulling force P exceeds the limiting friction, the resultant force accelerates the block.
– Once in motion, the frictional force decreases. The frictional force involved now is the kinetic friction.
– To maintain constant velocity, the pulling force P has to be decreased to the same magnitude as the frictional force (kinetic friction).
– The kinetic friction is independent of the relative velocity of the surfaces.
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