Question
A 200 N mass is supported by two wires that make angles of 54 degrees and 67 degrees with the ceiling.
Determine the tension in each wire.
Determine the tension in each wire.
Answers
Reiny
Make 2 diagrams, one called the position diagram, which shows the
actual situation, marking the angles at the ceiling and showing the hanging
mass of 200 N.
the other is a vector diagram where the sides of the triangle show the tension
I labelled mine ABC where angle A = 23°, angle B = 121° and angle C = 36°
with AC = 200
by the sine law:
AB/sin36 = 200/sin121
AB = 200sin36/sin121 = 137.146 N
BC/sin23 = 200/sin121
BC= 200sin23/sin121 = 91.168 N
check my arithmetic
actual situation, marking the angles at the ceiling and showing the hanging
mass of 200 N.
the other is a vector diagram where the sides of the triangle show the tension
I labelled mine ABC where angle A = 23°, angle B = 121° and angle C = 36°
with AC = 200
by the sine law:
AB/sin36 = 200/sin121
AB = 200sin36/sin121 = 137.146 N
BC/sin23 = 200/sin121
BC= 200sin23/sin121 = 91.168 N
check my arithmetic
henry2,
All angles are measured CCW from +x-axis.
T1*sin36+T2*sin157 = 200
Eq1: 0.59T1 + 0.39T2 = 200.
T1*cos36 = -(T2*cos157)
0.81T1 = -(-.92T2)
T1 = 1.14T2.
In Eq1, replace T1 with 1.14T2 and solve for T2:
0.59*1.14T2 + 0.39T2 = 200
0.67T2 + 0.39T2 = 200
1.06T2 = 200
T2 = 188.7 N.
T1 = 1.14*188.7 = 215 N.
T1*sin36+T2*sin157 = 200
Eq1: 0.59T1 + 0.39T2 = 200.
T1*cos36 = -(T2*cos157)
0.81T1 = -(-.92T2)
T1 = 1.14T2.
In Eq1, replace T1 with 1.14T2 and solve for T2:
0.59*1.14T2 + 0.39T2 = 200
0.67T2 + 0.39T2 = 200
1.06T2 = 200
T2 = 188.7 N.
T1 = 1.14*188.7 = 215 N.