ChemGeek16
Badges: 11
Rep:
?
#1
Report Thread starter 4 years ago
#1
Can someone talk me through these two questions?
1) A rod of a negligible weight and 0.75m long is pivoted at one end and a mass of 10kg is suspended from the other end. The rod is maintained in a horizontal position by a vertical string fastened to it 0.15m from the pivot. What is the tension in the string?

2) a uniform rod 5m long weighs 60N. If a piece of brass of weight 40N is attached at one end, how far from that end is the point at which the rod will balance?

Thanks in advance
0
reply
Lauren99
Badges: 2
Rep:
?
#2
Report 4 years ago
#2
A moment is the force multiplied by the perpendicular distance.

For the first question, you have 3 forces acting: Tension (T) acting upwards (we know the direction because we can assume that the system is in equilibrium and there's no horizontal forces acting), the weight (mg), and a reaction force (R) where the pivot it (acting upwards as there are no horizontal forces). You don't know the value of R so I would suggest taking the moments about the pivot, you'll be left with an equation where the only unknown is T, so you can solve it.

For the second, your forces are: 60N acting at the centre of the rod, 40N acting at one of the ends, and a reaction force at the pivot. Take the distance from the pivot to the 40N end (lets call this A) to be x. The 60N force acts 2.5m from A. Using this, we can work out that the perpendicular distance between the pivot point and the centre of mass of the rod is equal to 2.5 - x. Take your moments about the pivot, so you don't have to worry about the reaction force. This will give you an equation for x.

Hope this helps
0
reply
ChemGeek16
Badges: 11
Rep:
?
#3
Report Thread starter 4 years ago
#3
(Original post by Lauren99)
A moment is the force multiplied by the perpendicular distance.

For the first question, you have 3 forces acting: Tension (T) acting upwards (we know the direction because we can assume that the system is in equilibrium and there's no horizontal forces acting), the weight (mg), and a reaction force (R) where the pivot it (acting upwards as there are no horizontal forces). You don't know the value of R so I would suggest taking the moments about the pivot, you'll be left with an equation where the only unknown is T, so you can solve it.

For the second, your forces are: 60N acting at the centre of the rod, 40N acting at one of the ends, and a reaction force at the pivot. Take the distance from the pivot to the 40N end (lets call this A) to be x. The 60N force acts 2.5m from A. Using this, we can work out that the perpendicular distance between the pivot point and the centre of mass of the rod is equal to 2.5 - x. Take your moments about the pivot, so you don't have to worry about the reaction force. This will give you an equation for x.

Hope this helps
Thank you!! 😊
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you don't put your camera on in online lessons, why is that?

My teacher doesn't want us to (8)
19.05%
No one else does (16)
38.1%
I'm embarrassed about my background (4)
9.52%
I feel self-conscious showing my face (12)
28.57%
We don't use a video platform (1)
2.38%
I don't have a camera (0)
0%
Something else (tell us in the thread) (1)
2.38%

Watched Threads

View All
Latest
My Feed