What is the Angle Made by a Pendulum Using Parallel Lines?

[Priyadarshini]

Homework Statement

Homework Equations

The Attempt at a Solution

Using parallel lines I got the angle as theta.

 

[Suraj M]

Identify the forces on the pendulum bob through components of mg using theta

 

[Priyadarshini]

Identify the forces on the pendulum bob through components of mg using theta

The components of weight are mgcostheta and mgsintheta and there is tension acting in the string.

 

[Suraj M]

Could you draw a diagram to represent those forces on the bob,?

 

[Priyadarshini]

Could you draw a diagram to represent those forces on the bob,?

 

[Suraj M]

Since it's in free fall along the plane I think you should take a pseudo force along that line of motion($mg\sin\theta$)
Extend the length of the pendulum and label $\alpha$ I think you can proceed from there.

 

[haruspex]

View attachment 95005

What is providing this force N on the bob?

 

[Priyadarshini]

Since it's in free fall along the plane I think you should take a pseudo force along that line of motion($mg\sin\theta$)
Extend the length of the pendulum and label $\alpha$ I think you can proceed from there.

The tension in the string has components too. So T sin alpha=mg
Sin alpha = 1
So, alpha is 90.
Thank you!

 

[haruspex]

The tension in the string has components too. So T sin alpha=mg
Sin alpha = 1
So, alpha is 90.
Thank you!

Right answer, but I am baffled by your path to it.
I assume you are taking alpha as the angle between the string and the roof. If so, T sin alpha is not mg. And to perform your next step, you somehow had to have that T=mg. Where did that come from?

 

[Priyadarshini]

Right answer, but I am baffled by your path to it.
I assume you are taking alpha as the angle between the string and the roof. If so, T sin alpha is not mg. And to perform your next step, you somehow had to have that T=mg. Where did that come from?

Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.

 

[Priyadarshini]

Right answer, but I am baffled by your path to it.
I assume you are taking alpha as the angle between the string and the roof. If so, T sin alpha is not mg. And to perform your next step, you somehow had to have that T=mg. Where did that come from?

Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.

 

[haruspex]

Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.

Two forces act on the bob, the tension in the string and mg. You don't yet know what direction the tension is in. Also, the bob is accelerating, so the net force is not zero.
What you do know is that there is no acceleration perpendicular to the plane, so the forces must balance in that direction. But that still leaves you with two unknowns, T and alpha. So you need to use the known downplane acceleration of the system.

 

[Priyadarshini]

Two forces act on the bob, the tension in the string and mg. You don't yet know what direction the tension is in. Also, the bob is accelerating, so the net force is not zero.
What you do know is that there is no acceleration perpendicular to the plane, so the forces must balance in that direction. But that still leaves you with two unknowns, T and alpha. So you need to use the known downplane acceleration of the system.

But won't the vertical component of tension be T sin alpha anyway? And the vertical components need to be balanced. But then which force would I equate it to?

 

[haruspex]

But won't the vertical component of tension be T sin alpha anyway? And the vertical components need to be balanced. But then which force would I equate it to?

The component of the tension perpendicular to the roof will be T sin alpha, but that is not vertical.
The vertical components must balance if the acceleration has no vertical component, but it will have.

 

[Suraj M]

The component of gravitational force perpendicular to the inclined plane? Did you consider that? Relate that to the tension

 

[Priyadarshini]

The component of gravitational force perpendicular to the inclined plane? Did you consider that? Relate that to the tension

The component of the tension perpendicular to the roof will be T sin alpha, but that is not vertical.
The vertical components must balance if the acceleration has no vertical component, but it will have.

Can I do this:
mgsin theta + Tcos alpha - mgsin theta =0

mgsintheta is the horizontal weight component and the horizontal component of T acts in the same direction as it. The -mgsintheta is from the pseudo force that acts on the block.

So, Tcos alpha = 0
Cos alpha =0
Alpha = 90
]

 

[haruspex]

mgsintheta is the horizontal weight component and the horizontal component of T acts in the same direction as it. The -mgsintheta is from the pseudo force that acts on the block.

That works if you change all occurrences of 'horizontal' to ... what?