A 20.0 kg floodlight in a park is supported at the end of a horizontal beam of n

Question

A 20.0 kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole. A cable at an angle of 30.0 degrees with the beam helps support the light. Find (A) the vertical component of the force exerted by the pole on the beam. (B) the horizontal component of the force exerted by the pole on the beam. (C) the tension in the cable. Compute these values by computing torques around the junction between the cable and the beam at the right hand of the beam. I know that this can be done using newtons laws (by drawing a force diagram) but how do I solve this with torque?

Explanation / Answer

Frist you need to determine the vertical force created by the 20.0 kg floodlight. Since gravity is the only force acting on the floodlight this gives: F = m * a; a = g = -9.81m/s^2, m = 20.0kg F = 20.0kg * 9.81m/s^2 = 196.2kg*m/s^2 = 196.2N To determine the tension in the cable, you should draw a free body diagram and seperate the forces into x and y components. If you do this you will discover: For the vertical component: T * sin(30°) = Fy; T = cable tension, Fy = the vertical component of cable tension. If you look at the free body diagram and sum the forces in the vertical direction you will discover: SF(vertical) = 0 = -196.2N + Fy; this implies that Fy = 196.2N. The summation of forces in the vertical direction is equal to zero since the accleration of the system is also equal to zero. Since you know Fy, solve the equation above for T. This gives: T * sin(30°) = Fy T = Fy / sin(30°); plugging in the numbers gives: T = 196.2N / 0.5 = 392.4N = Tension in the cable. From the free body diagram you also know the horizontal component is: Fx = T * cos(30°); plugging in the known values gives: Fx = 392.4N * .866 = 339.81N; since you know the summation of the forces in the horizontal direction are equal to zero looking at the free body diagram wiill show: SFx = 0 = -Fx + Rp; Rp = reaction force at pole Fx = Rp = 339.81N = the horizontal forces exerted on the beam by the pole.

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