Resolving forces: Mass on a string

Click For Summary
A 50-gram mass hangs in equilibrium on a string, with a 0.3N force applied at a 30° angle to the horizontal. The confusion arises regarding the angle the string makes with the vertical, as it was initially thought to be 60°, but the correct approach involves considering the resultant of the gravitational and applied forces. To find the tension in the string, the horizontal and vertical components of the forces must be resolved, leading to different tension values. Clarification is needed on whether a free body diagram has been drawn to visualize the forces acting on the mass. Understanding these concepts is crucial for solving the problem accurately.
thebosonbreaker
Messages
32
Reaction score
5

Homework Statement


A mass of 50 grams hangs in equilibrium on a string. The mass is pulled aside and upwards by a force of 0.3N which makes an angle of 30° with the horizontal. Find the angle that the string makes with the vertical and the tension in the string.

Homework Equations


Body is in equilibrium, so I will need to resolve forces (tension and weight).

The Attempt at a Solution


For the first part of the question (find the angle that the string makes with the vertical):
I do not understand why this is not 60° - the string will move in the same direction as the 0.3N force applied to it, so if it makes an angle of 30° with the horizontal (the angle between 'horizontal'/'vertical' being a right angle), the angle made with the vertical must be: 180 - (90 + 30) = 60°. Could somebody please explain why this is not the case and how to find the angle?

For the second part of the question (finding the tension in the string):
Resolving horizontally I have that: Tcos30° = 0.3 [the horizontal component of tension equals the horizontal force applied]
Resolving vertically I have that: Tsin30° = mg = 0.05g
These both lead to different answers for T.

Could somebody clraify this problem? I'm having a bit of a hard time with it.
Thanks a lot in advance, your help will be greatly appreciated!
 
Physics news on Phys.org
thebosonbreaker said:

Homework Statement


A mass of 50 grams hangs in equilibrium on a string. The mass is pulled aside and upwards by a force of 0.3N which makes an angle of 30° with the horizontal. Find the angle that the string makes with the vertical and the tension in the string.

Homework Equations


Body is in equilibrium, so I will need to resolve forces (tension and weight).

The Attempt at a Solution


For the first part of the question (find the angle that the string makes with the vertical):
I do not understand why this is not 60° - the string will move in the same direction as the 0.3N force applied to it, so if it makes an angle of 30° with the horizontal (the angle between 'horizontal'/'vertical' being a right angle), the angle made with the vertical must be: 180 - (90 + 30) = 60°. Could somebody please explain why this is not the case and how to find the angle?

For the second part of the question (finding the tension in the string):
Resolving horizontally I have that: Tcos30° = 0.3 [the horizontal component of tension equals the horizontal force applied]
Resolving vertically I have that: Tsin30° = mg = 0.05g
These both lead to different answers for T.

Could somebody clraify this problem? I'm having a bit of a hard time with it.
Thanks a lot in advance, your help will be greatly appreciated!
In equilibrium, the string aligns in the direction of the resultant of gravity and the applied force.
 
If the object were heavy enough the applied force might only move it a very short distance. No where near 60 degrees.
 
The thread title mentions a spring. I don't see any mention of a spring in the problem itself. o_O

Edit: Does the thread title need correction?
 
Have you drawn a free body disgram showing the forces acting on the mass, or do you feel like you have advanced beyond the point where you need to use free body diagrams?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K