A hungry bear weighing 740 N walks out on a beam in an attempt to retrieve a bas

Question

A hungry bear weighing 740 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam (see figure below). The beam is uniform, weighs 200 N, and is 7.00 m long, and it is supported by a wire at an angle of = 60.0°. The basket weighs 80.0 N

(b) When the bear is at x = 0.80 m, find the tension in the wire supporting the beam and the components of the force exerted by the wall on the left end of the beam.

(c) If the wire can withstand a maximum tension of 875 N, what is the maximum distance the bear can walk before the wire breaks?
m

Explanation / Answer

(b) Because the beam is static, the counterclockwise torque about the support is zero. T is tension:

-740*0.8-200*3.5 - 80*7 + T*7*sin 60 degree = 0

T = (740*0.8 + 200*3.5 + 80*7 )/(7*sin 60 degree)

= 305.5 N

The sum of the horizontal components of the forces is zero.
Fx is the horizontal reaction:

Fx + Tcos( 120º ) = 0

Fx = -Tcos( 120º )

= -305.5 * (-0.4)

= 122.2 N towards the right

The sum of the vertical components of the forces is zero.
Fy is the horizontal reaction:

Fy - 740 N - 200 N - 80 N + T sin( 60 degree ) = 0

Fv -1020 N + 305.5*sin 60 degree = 0

Fv =  1020 N - 305.5*sin 60 degree

= 755.43 N upward

(c) The counterclockwise torque about the support is again zero:

-740*X - 200*3.5 - 80*7 + 875*7*sin 60 degree = 0

X = (200*3.5 + 80*7 - 875*7*sin 60 degree)/740

= 5.46 m
Answer

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