09.4 A restaurant is putting up a new sign. The sign weighs 80 pounds and is att

Question

09.4 A restaurant is putting up a new sign. The sign weighs 80 pounds and is attached to the end of a 50-pound horizontal beam of length 3.0 m. The beam is supported by a very light cable that is attached to the beam at a distance x from the wall, making an angle of ? = 30o with the beam. The cable is rated at 300 pounds, i.e., that is the maximum stress it can bear before breaking. What is the minimum distance x such that the cable does not break? At that distance, what is the force the wall exerts on the beam (magnitude and direction)?

Explanation / Answer

Here,

mass of sign , W1 = 80 pound

weight of beam , W2 = 50 pound

L = 3 m

theta = 30 degree

let the minimum distance is x

Balancing the moment about the wall support

W1 * L + W2 * L/2 - T * sin(30) * x = 0

80 * 3 + 50 * 3/2 - 300 * sin(30) * x = 0

x = 2.1 m

the minimum distance x such that the cable does not break is 2.1 m

-------------------

Now , for the reaction force from the wall

Fy = 80 + 50 - 300 * sin(30)

Fy = -20 N

Fx = 300 * cos(30) = 260 pound

force = sqrt(20^2 + 260^2)

Force = 260.7 pound

the force from the wall is 260.7 pound

theta = arctan(20/260)

theta = 4.4 degree

the direction of force is 4.4 degree above the horizontal

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