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Unbalanced force system results in accelerated motion.

We are now appropriately equipped to study motion under a system of forces, which is not a balanced force system. The resulting motion, in this situation, is an accelerated motion. The approach and methodology for studying motion with unbalanced force system are same as employed for the study of balanced force system.

There are large numbers of natural motions that fall under this category. One such motion involves motional mechanism that allows animals to move around – surprisingly using internal muscular force. We shall find out that this is not an exception to the laws of motion, rather an intelligent endeavor on the part of living beings that distinguish them from non-living counterparts.

There are many more such real time examples. A large numbers of real time examples of accelerated motion belong to the category of circular motion. We shall, however, save these circular motion cases for the study at a later stage.

Basic framework

The laws of motion are again the basic frame work for the study of body system under unbalanced force system. The treatment of the laws differs in only one respect to the one used for balanced force system. The net force, now, does not equate to zero, but to the product of mass and acceleration.

F = m a

In component form,

F x = m a x ; F y = = m a y ; F z = = m a z

Problem : A block of mass “m” is pulled by a string on a smooth horizontal surface with acceleration “a”. The string maintains an angle “θ” with horizontal. Find (i) tension in the string (ii) force applied by the surface on the block and (iii) force applied by the block on the surface.

Unbalanced force system

Solution : Here, we consider block as the body system.

Free body diagram of the block

The external forces on the block are (i) Tension in the string,T, (ii) weight of the block, mg, and (ii) Normal force applied by the surface,N.

Free body diagram

The block moves in x-direction with acceleration “a’. On the other hand, there is no motion involved in vertical i.e. y-direction as implied by the question. The forces in y-direction, therefore, constitute a balanced force system. Now,

F x = T cos θ = m a x T cos θ = m a T = m a cos θ

and

F y = N + T sin θ - m g = 0 N = m g - T sin θ

Putting values of tension, T, in the above equation,

N = m g - m a sin θ cos θ N = m g - m a tan θ

According to Newton’s third law, the surface is acted upon by the same magnitude of force that the surface applies on the block. Thus, force on surface is also N, acting downward.

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Note in the example above that vertical component of tension has reduced the normal force from its magnitude, when external force (T) is not applied. This highlights the important aspect of contact forces that they need not be equal to the weight of the body in contact. Equality of contact forces, however, is always maintained in accordance with Newton’s third law. This means that surface is also acted by a reduced normal force.

Composite mass system

A composite body system can be selected with elements having same acceleration. We must understand here that when we consider a composite body system then all internal forces should be neglected from our consideration. The external force causes a common acceleration to the composite body system. Clearly, this technique of combining bodies into a single composite body system is possible only when accelerations of the constituents are same.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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