Question

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.

Short answer

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

Step-by-step solution

Meaning of translational equilibrium

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 information

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 \({\theta _{\rm{o}}} = 15^\circ \).

Evaluation of the initial tension in the cord

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.

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

Evaluation of the force exerted by the bear

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 \({\rm{566}}\;{\rm{N}}\).