Question

A uniform beam weighing 1 N is used as a metre rule. The beam is supported by a pivot
at the 30 cm mark. A 12 N weight is hung at the 0 cm mark and a 2 N weight is hung at the
60 cm mark as shown in Figure 5.19.
O cm
50 cm 60 cm
100 cm
수
► Figure 5.19
30 cm
mark 1 N
12 N
2 N
W
What is the weight W of a mass hung at the 100 cm mark in order to balance the beam?
Please help! ​

Answer 1

Answer:

then you solve two torque one is anticlockwise and one is clock wise toque 1=force*perpendicular distance is equal to torque 2= force* perpendicular distance so tau = torque 1+ torque 2and give your answer

Answer 2

Answer:

W =  4 N (to balance the beam)

Explanation:

take moment at support Δ

∑M at Δ = 0 (note: clockwise is + and counter clockwise is - )

0 = W(100 - 30) + 2(60-30) + 1(50-30) - 12(30)

0 = W70 + 2(30) + 1(20) - 360

0 = W70 + 60 + 20 - 360

0 = W70 - 280

280 = W70

W = 280 / 70

W = 4 N (to balance the beam)

to take the reaction at support Δ

Δ = 12 + 1 + 2 + 4

Δ = 19 N

proof that W = 4

∑M at 12N = 0

0 = 19(30) - 1(50) - 2(60) - W(100)

0 = 570 - 50 - 120 - 100W

100W = 400

W = 400 / 100

W = 4 N -------- ok