PLEASE SHOW YOUR WORK. THANK YOU. The picture above shows a horizontal 3.5 m lon

Question

PLEASE SHOW YOUR WORK. THANK YOU.

The picture above shows a horizontal 3.5 m long, 40 kg beam of uniform density. The red X indicates where it is attached to the building by a hinge. The person on the beam has a mass of 70 kg, and is located0.91 m from the left edge. The right edge of the beam is attached by a cable to the building, making a 55-degree angle with the beam.

a) Determine the torque the beam exerts about the hinge.
(Watch the sign!... we call clockwise negative and counterclockwise positive.)

______ Newton-meters

(b) Determine the torque the person exerts about the hinge.
(Watch the sign!)
_____Newton-meters
(c) Determine the torque the cable exerts about the hinge.
_____Newton-meters
(d) Determine the magnitude of the force the cable exerts on the beam.
_____Newtons
(e) What is the X component of the force the hinge exerts on the beam?
____Newtons
(f) What is the Y component of the force the hinge exerts on the beam?
Newtons

Explanation / Answer

Let

m = 40 kg

M = 70 kg

L = 3.5 m

a) the torque the beam exerts about the hinge = -m*g*(L/2)*sin(90)

= -40*9.8*(3.5/2)*1

= -686 N.m

b) the torque the person exerts about the hinge = -M*g*d*sin(90)

= -70*9.8*0.91*1

= -624.26 N.m

c) the torque the cable exerts about the hinge. = 686 + 624.26

= + 1310.26 N.m

d) T*L*sin(55) = 1310.26

T = 1310.26/(L*sin(55))

= 1310.26/(3.5*sin(55))

= 457 N

e) Fx = Tx

= 457*cos(55)

= 262.1 N

f) Fnety = 0

Ty - Fy - m*g - M*g = 0

Fy = g*(m+M) - Ty

= 9.8*(40+70) - 457*sin(55)

= 703.6 N

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