Question

A hungry bear weighing 735 N walks out on a beam in an attempt
to retrieve a basket of goodies hanging at the end of the beam (see
the figure below). The beam is uniform, weighs 200 N, and is 5.00 m
long, and it is supported by a wire at an angle of *θ* =
60.0°. The basket weighs 80.0 N.

A horizontal plank is attached at the left end to a vertical
wall. A rod with a pulley at the right end extends a distance
*x* horizontally to the right from the wall above the plank.
A cable connected to the wall just above the rod goes around the
pulley then down and right to the end of the plank, making an acute
angle *θ* with the plank. A basket of goodies hangs down
from the right end of the plank and a bear stands on the plank
below the rod a distance *x* from the wall.

(a) Draw a force diagram for the beam. (Submit a file with a maximum size of 1 MB.)

(b) When the bear is at * x* = 1.04 m, find the
tension in the wire supporting the beam.

When the bear is at * x* = 1.04 m, find the
components of the force exerted by the wall on the left end of the
beam. (Assume the positive +

F_{x} |
= N |

F_{y} |
= N |

(c) If the wire can withstand a maximum tension of 800 N, what is
the maximum distance the bear can walk before the wire
breaks?

m

Answer #1

Okay, from the diagram description, I got this idea

take torque about left end

Tsin * L = 735 * 1.04 + 200 * 2.5 + 80 * 5

T sin 60 * 5 = 735 * 1.04 + 200 * 2.5 + 80 * 5

so,

**T = 384.37 N**

_____________________

**Fx = T cso 60 =192.2 N**

Fy = W + mg + Mg - Tsin 60

**Fy = 682.12 N**

______________________

use same equation as part (a) , this time we need to solve the distance of bear from left end

800 sin 60 *5 = 735 * x + 200 * 2.5 + 80 * 5

solve for x, I got

x = 3.488 m

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