Consider the following situations:

 You and a friend are sitting on chairs with rollers.  You both have your feet off the ground and with your feet you push against your friend's chair.  Your friend starts moving away from you and you start moving away from your friend.  You both are accelerating.  The direction of your acceleration is opposite to the direction of your friend's acceleration. You both stretch out your hands and you grab your friend and pull him towards you.  He accelerates towards you and you accelerate towards him.  Again you both are accelerating. What is happening is a consequence of Newton's third law. For every force that an object exerts on a second object, there is a force equal in magnitude but opposite in direction exerted by the second object on the first object.   Forces are the result of interactions.

You pull on me, I pull on you.  Assume no other forces are present.  I accelerate towards you.  My acceleration has the magnitude of the force divided by my mass.  You accelerate towards me.  Your acceleration has the magnitude of the force divided by your mass.

Question:

You push against the table, the table pushes against you.  You accelerate away from the table.
Why is the table not accelerating away from you?

 If the table is on rollers, it will accelerate away from you.  If the table is very massive, the magnitude of its acceleration will be much smaller than the magnitude of your acceleration.  If it is not on rollers, then the force of static friction is also acting on it, and the vector sum of the two forces is zero.

Standing on a very slippery surface or on a skateboard, you throw a heavy object away from you in a northerly direction.  To do this, you have to exert a force on the object in the northerly direction.  The object exerts a force on you, which has equal magnitude but points towards the south.  You accelerate southward.

Newton's third law is also called the law of action and reaction.  For every action force, there exists a reaction force, equal in magnitude and opposite in direction.  The action and reaction force always act on different objects.

#### Free body diagrams

To find the net force on an object, all the vector forces that act on the object have to be added.  We use free-body diagrams to help us with this task.  Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting on an object in a given situation.  The object is represented by a box.  The forces are represented by arrows.  The length of the arrow in a free-body diagram is proportional the magnitude of the force.  The direction of the arrow gives the direction of the force.  Each force arrow in the diagram is labeled.  All forces, which act on the object, must be represented in the free-body diagram.  A free body diagram only includes the forces that act on the object, not the forces the object itself exerts on other objects.

#### Examples:

 A free-body diagram for a freely falling ball: Neglecting air friction, the only force acting on the ball is gravity. A free-body diagram for a ball resting on the ground: Gravity is acting downward.  The ball is at rest.  The ground must exert a force equal in magnitude and opposite in direction on the ball.  This force is called the normal force, n, since it is normal to the surface.How would you draw a free body diagram for the ground? A free-body diagram for a mass on an inclined plane: Gravity acts downward.  The component of Fg perpendicular to the surface is cancelled out by the normal force the surface exerts on the mass.  The mass does not accelerate in the direction perpendicular to the surface.  The component of Fg parallel to the surface causes the mass to accelerate in that direction.

#### Problem:

While a football is in flight, what forces act on it?   What are the action and reaction pairs while the football is being kicked and while it is in flight?

Solution:
 While in flight, gravity and air friction act on the ball.When the ball is being kicked, the earth pulls on the ball downward (gravity), the ball pulls on the earth upward (gravity).  The foot pushes the ball forward, the ball pushes the foot backward.When the ball is in flight, the earth pulls on the ball downward (gravity), the ball pulls on the earth upward (gravity).  The ball pushes the air forward and the air pushes the ball backward.

#### Tension

Assume your team and an opposing team are pulling on a rope in opposite directions.  The rope starts accelerating in the direction of the opposing team.
What is the net force on the rope?

 It is the vector sum of the force of gravity and the forces exerted by the two teams.

Which team is exerting the greater force?

 The opposing team, since the rope starts accelerating in their direction.

Is there a net force on the rope if the rope is not accelerating, even so each team is pulling as hard as it can?

 No, there is no net force.  However, there is tension in the rope.  Tension results from different forces acting on different parts of the body.  Tension can break things.  A pure force, i.e. the same force acting on all parts of the body, cannot break things.If instead of pulling on the rope the two teams push on a heavy rock, but the rock does not move, then again the net force on the rock is zero.  However, now the rock is under compression.

The tension T is a scalar.  It is defined as the magnitude of the force with which the stretched rope pulls on whatever it is attached to.  This force is often denoted by T.  The direction of T depends on which end of the rope is being considered.

The net force acting on an object is always the vector sum of all forces acting on an object.  The object's acceleration is in the direction of this net force and has magnitude a = (net force/object's mass).

#### Problem:

A 15-lb block rests on the floor.
(a)  What force does the floor exert on the block?
(b)  If a rope is tied to the block and run vertically over a pulley and the other end is attached to a free-hanging 10-lb weight, what is the force exerted by the floor on the 15-lb block?
(c)  If we replace the 10-lb weight in part (b) with a 20-lb weight, what is the force exerted by the floor on the 15-lb block?

Solution:
 (a) The block is at rest the net force is zero.  The weight of the block is 15 lb downward.  The floor must therefore exert a force 15 lb upward. (b) The block is still at rest.  The weight is 15 lb downward, the rope pulls with 10 lb upward due to the tension in the rope.  The floor must therefore exert a force 5 lb upward. (c) The weight is 15 lb downward, the rope pulls with 20 lb upward due to the tension in the rope.  The block will accelerate upward, since the net force is upward.  The block is not pushing on the ground, and the ground is not pushing back.  The ground exerts zero force on the block.

Problem:

 (a)  Find the tension in each of the cords of the system shown in the figure below.  Neglect the masses of the cords. (b)  Find the tension in each of the cords of the system shown in the figure below.  Neglect the masses of the cords.
Solution:
 (a) The mass is not accelerating, it is in equilibrium.  The net force on the mass is zero.  We therefore have T3 = mg = 49 N.The knot in the cable is in equilibrium.  The net force on the knot is zero.  Fx = 0, Fy = 0. ∑Fix = 0, ∑Fiy = 0.  Adding the x-components of all forces in the diagram we obtain Fx = -T1cos40o + T2cos50o = 0 , or T2 = T1cos40o/cos50o = .192 T1.Adding the y-components we obtain Fy= T1sin40o + T2sin50o - T3 = 0, or 0.643 T1 + 1.192 T1*0.766 = 49 N, 1.556T1 = 49 N, T1 = 31.5 N, T2 = 37.54 N. (b) The mass is not accelerating, it is in equilibrium.  The net force on the mass is zero.  We therefore have T3 = mg = 98 N.The knot in the cable is in equilibrium.  The net force on the knot is zero.  Fx = 0, Fy = 0. ∑Fix = 0, ∑Fiy = 0.  Adding the x-components of all forces in the diagram we obtain Fx = -T1cos60o + T2 = 0, or T2 = T1cos60o = 0.5T1.Adding the y-components we obtain Fy = T1sin60o - T3 = 0, or 0.866 T1 = 98 N, T1 = 113.2 N, T2 =  56.6 N.

Problem:

 A bag of cement hangs from three wires as shown in the figure.  If the system is in equilibrium,(a)  show that T1 = w cosθ2/sin(θ1 + θ2).(b)  Given that w = 325 N, θ1 = 10o and θ2 = 25o, find the tensions T1, T2, and T3 in the wires.
Solution:
 (a) The mass is not accelerating, it is in equilibrium.  The net force on the mass is zero.  We therefore have T3 = w.The knot in the cable is in equilibrium.  The net force on the knot is zero.  Fx = 0, Fy=  0.   ∑Fix = 0, ∑Fiy = 0.  Adding the x-components of all forces in the diagram we obtain Fx = -T1cosθ1 + T2cosθ2 = 0, or T2 = T1cosθ1/cosθ2.Adding the y-components we obtain Fy = T1sinθ1 + T2sinθ2 - T3 = 0, or T1sinθ1 + T1sinθ2 cosθ1/cosθ2 = w.  We rewrite this equation as T1(sinθ1cosθ2 + sinθ2cosθ1) = wcosθ2.sinθ1cosθ2 + sinθ2cosθ1 = sin(θ1 + θ2).T1 = wcosθ2/sin(θ1 + θ2). (b) T1 = (325 N)(0.906/0.574) = 513.5 N, T2 = T1(0.985/0.906) = 558 N, T3 = 325 N.

Problem:

Draw a free body diagram of a block that slides down a frictionless plane having an inclination of θ =15o.  If the block starts from rest at the top and the length of the incline is 2 m, find
(a)  the acceleration of the block and
(b)  its speed when it reaches the bottom of the incline.

Solution:
 (a) Please refer to the free-body diagram for a mass on an inclined plane.The acceleration of the block has magnitude gsinθ and its direction is parallel to the surface of the inclined plane, downhill.  We have a = 9.8(m/s2)*0.259 = 2.54 m/s2. (b) Choosing our x-axis parallel to the surface of the inclined plane and using the kinematic equationvxf2 = vxi2 + 2ax(xf-xi) we find vxf2 = 2*2.54 (m/s2)*2 m =10.15 (m/s)2. vxf = 3.2 m/s is the speed of the block when it reaches the bottom of the incline.

#### Measuring forces

Many objects stretch or contract when under tension or compression.  As long as the change in length compared to their equilibrium length is small, the change in length is often proportional to the applied force on either end.  If this is the case, we say that the object obeys Hooke's law, F = k∆l, where ∆l denotes the change in length.  For most objects that obey Hooke's law k is very small and ∆l is hard to measure, but for springs k is much larger and ∆l can easily be measured.  Springs can therefore be used to measure the magnitude of forces.  After they have been calibrated and k has been found, all we have to do is measure the change in length and multiply by k to obtain the magnitude of the force.