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Worked Examples · Example 4.7

Q.Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train's floor is 0.150.15.

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For the box to remain stationary relative to the accelerating train, static friction must provide the necessary force, up to its limiting value. The maximum acceleration is 1.5 m/s2\boxed{1.5\ \text{m/s}^2}.

When a box is placed on the floor of a train, and the train accelerates, the box tends to resist this change in motion due to its inertia. If there were no friction, the box would slide backward relative to the train. Static friction acts to prevent this relative motion.

For the box to remain stationary relative to the train, it must accelerate with the train. This means the static friction force acting on the box must be precisely what is needed to give the box the same acceleration as the train.

Static friction is self-adjusting: it supplies exactly the force needed to prevent relative motion, up to a maximum value μsN\mu_s N. If the required force exceeds this maximum, the box slides. The maximum train acceleration the box can withstand without sliding occurs when static friction reaches this maximum.

  1. Identify the forces acting on the box.

    • Gravity (mgmg): downward.
    • Normal force (NN): upward, from the train's floor.
    • Static friction (fsf_s): horizontal. Since the train accelerates forward, the box tends to lag behind relative to the train, so friction acts forward, in the direction of the train's acceleration, to hold the box with it.
  2. Vertical equilibrium.

    The box does not accelerate vertically:

N−mg=0  ⟹  N=mgN - mg = 0 \implies N = mg

  1. Maximum static friction.

fs,max⁡=μsN=μsmgf_{s,\max} = \mu_s N = \mu_s mg

  1. Horizontal Newton's second law, at the verge of slipping. …

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