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Friday, 26 August 2011

Physics GRE - #25

A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally until the horizontal force is 10 N. The box is then in equilibrium as shown below.

The angle that the rope makes with the vertical is closest to:

  1. $\arctan0.5$
  2. $\arcsin0.5$
  3. $\arctan2.0$
  4. $\arcsin2.0$
  5. $45^\circ$

Solution :

Choice 1 is the answer.

Since the block is in equilibrium, we must have $\vec{F}+\vec{F_g}=\vec{T}$ as shown below.

Hence $\theta=\arctan\frac{|\vec{F}|}{|\vec{F_g}|}=\arctan\frac{10}{2g}\approx\arctan\frac{10}{20}=\arctan0.5$.


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