Reaction Force Problems and Laws of Motion

In previous articles we discussed newton’s laws of motion and how they apply to life-like situations. In this article we’ll be using formulas/equations from the laws of motion and the laws themselves to solve various problems involving reaction force. By the end of this tutorial you should be able to:

  1. Find the reaction force in a system
  2. Apply newton’s laws of motion to various questions
  3. Find missing variables such as acceleration when given values for other variables.


Problem #1
A body of mass 7kg rest on a floor lift. Calculate the reaction force R exerted by the floor of the lift on the body if:

     a) The lift has an upward acceleration of 2m/s2.
     b) The lift has a downward acceleration of 3m/s2.
     c) The lift is moving with constant velocity.

Answers
The first thing that needs to be done is to draw a diagram to represent the information in the question, doing this makes your calculations allot easier and reduces risk of errors. In questions like these where there’s a mass there is always a weight acting downward from the body; so the first thing to be drawn on the diagram is the weight (W) acting downward. Next is the reaction force(R), the reaction force always acts perpendicularly to the body. When done your diagram should look something like this:



     a) This part of the question asked for the reaction force when the lift has an upward acceleration of 2m/s2.




When forces are moving in opposite directions you must subtract the smaller from the larger. In this question the reaction force is larger than the weight because the lift is moving in the upward direction.
We’ll use the formula derived from newton’s laws of motion:

Force = Mass x Acceleration
Your answer should know look something like this:

F = Mass × acceleration
R – W = Mass × acceleration
R – 70 = 7 x 2
R = 14 + 70
R = 84N


 Note: This is how the weight was found
Weight = mass x gravity
Weight = 7 x 10
Weight = 70N
     b) This part of the question is very similar to the one previously done. The question asked for the reaction force when given a downward acceleration of 3m/s2. In part “a” of the question the reaction force was larger because the lift was moving in an upward direction; therefore it also follows that the opposite should happen in this part of the question since the lift is now moving in the downward direction. Therefore the weight is larger than the reaction force.



F = mass x acceleration
70 – R = Mass x acceleration
70 – R= 7 x 3
70 – R = 21
70 – 21 = R
49N = R

     c) In this part of the problem you’re required to find the reaction force given that the lift is moving at a constant velocity. Since it is moving at constant velocity acceleration would be zero and no motion is taking place.

F = 0
R – W = 0
R – 70 = 0
R = 70N


Note:  
Here the reaction force is equal to the weight

Problem # 2
A body of mass 5kg is pulled up a smooth plane inclined at 300 to the horizontal by a force of 40N acting parallel to the plane. Calculate the acceleration of the body and the force exerted on it by the plane.



Answers
Before we begin calculating some other components must be added to the diagram. Remember in the previous question we said that where ever there’s a mass there is weight acting downwards. Also the weight must be resolved/split into its perpendicular components with respect to the plane. One of the components of weight is parallel to the plane while the other is perpendicular to the plane. Finally we must now add the reaction force. Remember that the reaction force must be perpendicular to the plane of the body.


You may have noticed that another angle was added to the diagram, 600, this was needed in order to find the perpendicular components of the weight. If you look at the diagram closely you’ll see that the line for the weight cuts the base of the triangle forming a 900 angle. Therefore the 60 degrees was found by subtracting the sum of the 90 degrees and 30 degrees from 180 giving 60 degrees. Because the sum of all angles in a triangle add up to 180.

Now that we have labeled our diagram we can now begin answering the question. The first part asked for the acceleration of the body.

Using newton’s second law
F = Mass x Acceleration (a)
40 – W x cos60 = 5 x a
40 – 50 x cos60 = 5 x a
40 – 50(0.5) = 5 x a
40 – 25 = 5 x a
15 = 5 x a
15/5 = a
3m/s2 = a

Note:

Remember that forces in opposite direction must be subtracted.
The next part of the question asked for the force exerted on the body by the plane, the force exerted is the reaction force. No motion is taking place in this direction therefore:

Using Newton’s first law:
F = 0
R – W x Sin60 = 0
R = W x Sin60
R = 50(0.8660)
R = 43.30N

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2 Response to Reaction Force Problems and Laws of Motion

February 24, 2012 at 4:03 PM

I,m a physics teacher and I really liked this post. However I suggest you make a post to discuss Newton's third law of motion seperatly

February 25, 2012 at 10:26 AM

Thanks for the tip aboabdullah

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