A street lamp weighs 190 N. It is supported by two wires that form an angle of 140° with each other?

Question

A street lamp weighs 190 N. It is supported by two wires that form an angle of 140° with each other. The tensions in the wires are equal.
(a) What is the tension in each wire?
____ N
(b) If the angle between the wires is reduced to 110°, what is the tension in each wire?
_____N

Answer 1

a) 190N=2Tcos 70
T=277.8 N

b)190 N=2Tcos55
T=165.6 N

*pretty sure*

Draw a free body diagram showing each of the forces. The weight of the lamp is in the negative y direction (270 degrees). One cable is located at 20 degrees and the other at 160 degrees (for part b, 35 degrees and 145 degrees). Resolve the forces of the cables into x and y components. Then balance the y components of the cable forces with the weight of the lamp. The y component of the cable forces for part a is Tcos70, but you have two of them so it is 2Tcos70, which is equal to the weight of the lamp, 190 N.

Answer 2

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)^2]] = 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.

Answer 3

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

Answer 4

Answer to (a) is 142.11N approximately. You have to resolve the forces. It is assume that the lamp is in equilibrium
For (b) it is 4067.474N
It is actually = Half the weight upon cos of half the angle.

Answer 5

a) f{y] = 0
2(t)cos(70) = 190
t = 190/(2cos(70))= 277.76 N each
b)f[y] = 0
2tcos(55) = 190
t = 190/(2cos(55)) = 165.63 N

Answer 6

(a) 95N
(b) it is also 95N because the angle only is reduced and not the weights.