Today we are going to come up with new task, that is about ”The 4.6 kg, uniform, horizontal rod in (Figure 1) is seen from the side. Part A…“. We have found the solution to this question, along with interesting information related to it, thanks to its novelty and interest. This will help you to develop a more research-oriented mindset, and make it easier for you to answer these types of questions. This is where we will focus to gain this useful and new knowledge!

Question: “The 4.6 kg, uniform, horizontal rod in (Figure 1) is seen from the side. Part A…”

Answer: “The 4.6 kg, uniform, horizontal rod in (Figure 1) is seen from the side. Part A…”

Calculate the Weight of Rod = mg
w = 4.6 * 9.81
w = 45.13 N

Calculate the Centre of gravity of Left side :
CGl = 25/2 = 12.5 from pivot

Calculate the Weight of left section :
Wl = 45.13*25/100 = 45.13*1/4
Wl = 11.28 N

Write the equation for torque on left side.

Torque = Force*Distance
Tl = 11.28*12.5
Tl = 141 N.cm will be anticlockwise

Calculate the Centre of gravity of Right side :
CGr = 75/2 = 37.5 from pivot

Calculate the Weight of right section :
Wr = 45.13 * 75/100 = 45.13*3/4
Wr = 33.85 N

Write the equation for torque on left side.

Torque = Force*Distance
Tr = 33.85 * 37.5
Tr = 1269.28 N.cm will be clockwise

Tnet = Tr – Tl
= 1269.28- 141= 1128.38 N.cm = 11.28 N.m

The positive sign of the net torque indicates the clockwise direction.

Relevant knowledge

The rotational equivalent to force is called torque. Torque is the ability of a force or object to cause rotational motion about a particular axis of rotation. It is mathematically calculated as the vector product between the position vector and the force vector, i.e.

Conclusion

This is how you solve this type of task “The 4.6 kg, uniform, horizontal rod in (Figure 1) is seen from the side. Part A…”. Please leave a message below if you have any questions or comments.

Thanks for reading!!!