Plotting the graph of Friction Force vs. time (Laws of motion)

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SUMMARY

The discussion focuses on the dynamics of two blocks subjected to friction forces and tension in a string. The maximum static friction for the 1 kg block is calculated as 5N, while for the 3 kg block, it is 15N. The tension in the string becomes significant when the applied force exceeds 5N, leading to different accelerations for each block. The challenge lies in plotting the graph of friction force (f2) against time (t) after the tension is established, as the relationship between tension and acceleration remains undefined without additional information.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Knowledge of static and kinetic friction coefficients
  • Familiarity with free body diagrams (F.B.D)
  • Basic principles of tension in inextensible strings
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  • Explore the concept of tension in systems with multiple blocks
  • Learn about graphing force versus time in dynamic systems
  • Study the effects of varying applied forces on block motion
  • Investigate the role of friction in multi-body dynamics
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SpectraPhy09
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Homework Statement
Coefficient of friction between all the surfaces in 0.5. Which diagram would describe the friction between the ground and the 2Kg Block when f = t N is applied on the 2kg Block? It is given that the string is not taut initially.(Plz check the the diagram in the image attached). take g= 10m/s^2
Relevant Equations
F = ma
F(friction) = μmg
F.B.D Of first block
(I have shown only the horizontal Forces)
F.B.D 1.png

f1(max) = μ (1kg)(g) = 0.5 * 10 = 5N
F.B.D Of the second Block
F.B.D 2.png

f2(max) = μ (3kg)(g) = 15N

Now the string will become taut and the tension will start acting when f = t = 5N
But for 0<f<5N there will be no motion between the 1 kg and the 2 kg block so the f2=0
but when f > 5N the string will become taut and the friction the Tension will start will acting now suppose the acceleration of the 1 kg block in left direction is a1 then,
(1)a1 = f - T - f1
and suppose the acceleration od the 2kg block is a2 then,
(2)a2 = f1 - f2 - T
But how will be plot the graph between f2 and t after t=5 since we do not know the T ension neither the acceleration of the block.
Also in the option for 0<t<5 there is different graph that mine.
Plz can someone help me how to solve this problem ahead?
 

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SpectraPhy09 said:
But for 0<f<5N there will be no motion between the 1 kg and the 2 kg block so the f2=0
If f2=0 and the string is not taut, what stops the pair of blocks moving left?
SpectraPhy09 said:
suppose the acceleration od the 2kg block is a2
If the 2kg block is moving, what is the state of the string? Assume it is inextensible.
SpectraPhy09 said:
(2)a2 = f1 - f2 - T
You have drawn f1 and f2 acting in the same direction on the 2kg block.
 

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