A hungry 603 N bear walks out on a beam in an attempt to retrieve some “goodies”

Question

A hungry 603 N bear walks out on a beam in an attempt to retrieve some “goodies” hanging at the end. The beam is uniform, weighs 304 N, and is 7.57 m long; the goodies weigh 88.7 N.

a. When the bear is 1.7 m from the hinged end, find the tension in the wire. Answer in units of N.

b. Find the vertical component of the reaction force at the hinge. Answer in units of N.

c. Find the horizontal component of the reaction force at the hinge. Answer in units of N.

d. If the wire can withstand a maximum tension of 437 N, what is the maximum distance the bear can walk before the wire breaks? Answer in units of m.

Explanation / Answer

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

-( 603N )( 1.7m ) -( 304N )( 3.785m ) -( 88.7N )( 7.57m ) + T( 7.57m )sin( 61º ) = 0
-2847.19Nm + T( 6.6208m ) = 0
T = 2847.19Nm / 6.6208m
T = 430.03N

b) The sum of the vertical components of the forces is zero.
Rv is the horizontal reaction:

Rv -603N -304N -88.7N + Tsin( 61º ) = 0
Rv -995.7N + ( 430.03N )( 0.8746 ) = 0
Rv -995.7N + 376.10N = 0
Rv = 995.7N - 376.10N
Rv = +619.6N ( upward )


c) The sum of the horizontal components of the forces is zero.
Rh is the horizontal reaction:

Rh + Tcos( 122º ) = 0
Rh = -Tcos( 122º )
Rh = -( 430.03N )( -0.5299 )
Rh = +227.8N ( toward the right )

d) The counterclockwise torque about the support is again zero:

-( 603N )( X ) -( 304N )( 3.785m ) -( 88.7N )( 7.57m ) + ( 437N ) ( 7.57m )sin( 61º ) = 0
-( 603N )( X ) -1822.099Nm+ 2893.32Nm = 0

-( 603N )( X ) +1071.22 = 0
X = 1071.22Nm / 603N
X = 1.776 m

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