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Q70GP

Expert-verifiedFound in: Page 230

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**A 23.0-kg backpack is suspended midway between two trees by a light cord as in Fig. 9–51. A bear grabs the backpack and pulls vertically downward with a constant force, so that each section of cord makes an angle of 27° below the horizontal. Initially, without the bear pulling, the angle was 15°; the tension in the cord with the bear pulling is double what it was when he was not. Calculate the force the bear is exerting on the backpack.**

** **

The force that the bear is exerting on the backpack is \({\rm{566}}\;{\rm{N}}\).

**A body is supposed to be in translational equilibrium when the summation of all the forces acting on the object is equivalent to zero.**

Given data:

The mass of the backpack is \(m = 23.0\;{\rm{kg}}\).

The final angle made by the cord with the horizontal is \(\theta = 27^\circ \).

The initial angle made by the cord with the horizontal is \({\theta _{\rm{o}}} = 15^\circ \).

The free-body diagram for the initial condition of the system can be drawn as follows:

Here, \({F_{{{\rm{T}}_{\rm{o}}}}}\) is the initial tension in the cord.

Apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\2{F_{{{\rm{T}}_{\rm{o}}}}}\sin {\theta _{\rm{o}}} - mg = 0\\2{F_{{{\rm{T}}_{\rm{o}}}}}\sin {\theta _{\rm{o}}} = mg\\{F_{{{\rm{T}}_{\rm{o}}}}} = \frac{{mg}}{{2\sin {\theta _{\rm{o}}}}}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}{F_{{{\rm{T}}_{\rm{o}}}}} = \frac{{\left( {23\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)}}{{2\sin \left( {15^\circ } \right)}}\\{F_{{{\rm{T}}_{\rm{o}}}}} = 435.4\;{\rm{N}}\end{array}\)

The tension in the cord with the bear pulling is doubled. Therefore, the relation between the initial tension and final tension can be written as follows:

\({F_{\rm{T}}} = 2{F_{{{\rm{T}}_{\rm{o}}}}}\)

The free-body diagram for the final condition of the system can be drawn as follows:

Here, \({F_{\rm{T}}}\) is the final tension in the cord, and \({F_{{\rm{Bear}}}}\) is the force exerted by the bear.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\2{F_{\rm{T}}}\sin \theta - mg - {F_{{\rm{Bear}}}} = 0\\2\left( {2{F_{{{\rm{T}}_{\rm{o}}}}}} \right)\sin \theta - mg - {F_{{\rm{Bear}}}} = 0\\{F_{{\rm{Bear}}}} = 2\left( {2{F_{{{\rm{T}}_{\rm{o}}}}}} \right)\sin \theta - mg\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{l}{F_{{\rm{Bear}}}} = 2\left[ {2 \times \left( {435.4\;{\rm{N}}} \right)} \right]\sin \left( {27^\circ } \right) - \left( {23\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\\{F_{{\rm{Bear}}}} = 565.26\;{\rm{N}} \approx {\rm{566}}\;{\rm{N}}\end{array}\)

Thus, the force that the bear is exerting on the backpack is \({\rm{566}}\;{\rm{N}}\).

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