A Street Lamp Weighs 190 N. It Is Supported by Two Wires That Form an Angle of 140 with Each Other?

a) Summation of forces on the y-axis T1 sin20 T2 sin20 = 190 Summation of forces along the x-axis T1 cos20 = T2 cos20 T1 = T2 a) T1 = 190/(2 sin20) = 277.76 N T2 = 277.76 N ----------these are the tension of the wires. b) If the angle between is reduced to 110 degrees. Then the tension will be, T1 =T2 = 190/( 2 sin 35) = 165.63 N

1. A street lamp weighs 110 N. It is supported by two wires that form an angle of 100Â° with each other. The tensi

Resolving forces vertically . Two angles make 100 degress with each other . make angle of 100/2 = 50 Degress with the vertical Tcos50Tcos50=110 T=85.5 N. b) Now they make an angle 0f 70/2=35 with the vertical Tcos35Tcos35=110 T=67.1 N.

2. A street lamp weighs 150 N. It is supported by two wires that form an angle of 120 degrees with each other..?

Any system...let me repeat...any system that is not accelerating (or decelerating) has balanced forces acting on it so that the net force (f) is exactly zero. This results because f = ma = 0 if and only if a = 0 for a mass m. OK your weight W = 150 N is not accelerating, in fact it is static; so there is no motion whatsoever. So f = 0 = ma for the system, which includes the weight and the two wires. This means that something is canceling out the weight because f = ma = W - F = 0; where F is that "something" canceling the weight (W). The only "something" you've defined for your system is the forces pulling up by the attached wires. Since there are two wires, each one has a share of that upward force. By "upward" we mean vertical force acting in a direction opposite of the weight (W). As there are 120 deg between the two wires, the vertical force for both wires is W cos(60) = Fv and for each individual wire, it's Fv/2 = (W/2) cos(60). Similarly, the horizontal force on each wire is Fh = (W/2) sin(60). Since the vertical and horizontal forces form a right angle for each wire, the force along each wire, the tension, is T = sqrt(Fv^2 Fh^2). And, by substitution, T = sqrt((W/2)^2 cos(60)^2 (W/2)^2 sin(60)^2) = sqrt[(W/2)^2 [sin(60)^2 cos(60)^] = sqrt((W/2)^2) = W/2. a. Thus, the tension (T) in each wire is one half the weight of the street lamp = W/2. b. Following the same line or reasoning used to find the tension at 120 deg, we see that the angle makes no difference because sin(deg)^2 con(deg)^2 = 1.0 no matter what deg is. Tension, therefore, will still be W/2 in each wire with 90 deg between them. The important thing to remember is that, if the system is not accelerating or decelerating, the net force on it is zero. So all the forces acting on it will cancel out.

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