A 6 kg lamp is suspended by two light wires. The first wire is horizontal and is

Question

A 6 kg lamp is suspended by two light wires. The first wire is horizontal and is attached to the wall. The second wire is attached to the ceiling and is at an angle 42.3° with respect to the horizontal as shown in the figure below.

What is the tension in the horizontal wire attached to the wall?

Main Menu Contents Grades Reservations Course Contents » » CORRECTION EXAMQuestion 12 ·O Timer Notes à Evaluate-Feedback-Print Ang kg lamp is suspended lDy two light wires The fret wire ks orizonta and is ttched to the wal. The second wire is attached to the celling A 6 kg lamp is suspended by two light wires. The first wire is horizontal and is attached to the wall. The second wire is attached to the ceiling and is at an angle 42.3° with respect to the horizontal as shown in the fiqure below and is at ah anale 42.3° with respect to the horizontal as shown in the figure below What is the tension in the horizontal wire attached to the wall? Submit Answer Tries 0/20 Threaded Vie ological View Other Vie iew _Other View My general preferences on what is marked as NEW

Explanation / Answer

Weight mg is acting vertically downwards. This weight is supported by two wires and therefore they must have tension T acting along the length of the wire in the upward direction.

Let T be the tension in the left wire which is at angle 42.3 degrees from the ceiling and T' be the tension in the right wire which is horizontal.

Now since the weight is not moving in the horizontal direction, the horizontal components of tension in both the wires must be balanced.

or Tcos42.3 - T' = 0 ...................................(1)

therefore, T' = Tcos42.3

Now the weight is suspended with the wires such that it has no motion in the y direction either. This means that the sum of vertical component of the tension in both wires must balance the weight mg of the lamp.

Now since the right wire is at right angle to the vertical therefore its component of tension in the vertical direction must be zero. Thus only the vertical component of the left wire keeps the lamp from falling.

therefore, Tsin42.3 = mg .........................................(2)

or Tsin42.3 = 6x9.8

or T = 87.36 N

Now from equation 1 we know that Tcos42.3 = T'

therefore T' = 87.36cos42.3 = 64.62N

this is the tension in the horizontal wire.

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