Solving Physics Exam Problems: Quadratic and Torque Examples

[donotremember]

I just wrote a physics exam today and I got stumped on a couple problems that have been bugging me, and I should probably find out how they are done the right way, or at least how to set them up just in case I use them again and I don't see any examples in my textbook:

The first I'm pretty sure involves a quadratic. It involves finding the angle you need to tilt a gun up in order to hit a target a distance d away if the target is on the same level as you. Only the Vi from the gun (without components) and the distance away are given.


The second is a torque question I think. Determine if a glass will tip over when a horizontal force is applied to the top rim and the coefficient of static friction is 0.2. The glass weighs w has height h and has diameter d.

 

[Doc Al]

Show what you did and where you got stuck.

 

[donotremember]

Show what you did and where you got stuck.

The distance equations for each component are

Given Vi

(1) Viy = Vi*sin(Θ) -1/2 (9.81)t^2

(2) Vix = Vi*cos(Θ) *t

I tried stating that

t = Vix/(Vi*cos(Θ))

and putting that in equation (1) for t and I get lost after I end up with sin(Θ) and cos(Θ)

I can't find examples of this problem in by textbook or notes so it's possible there is a much better approach.For the second problem the frictional force is just w(0.2) and acts in the direction resisting motion. The force that acts on the top rim acts at a distance to a pivot (I think) Possibly at the center of the glass.

 

[LowlyPion]

The distance equations for each component are

Given Vi

(1) Viy = Vi*sin(\theta) -1/2 (9.81)t^2

(2) Vix = Vi*cos(\theta) *t

I tried stating that

t = Vix/(Vi*cos(\theta))

and putting that in equation (1) for t and I get lost after I end up with sin(\theta) and cos(\theta)

I can't find examples of this problem in by textbook or notes so it's possible there is a much better approach.

That's the right approach.

You should end up with a term in sinθcosθ . That's what you want, because then you can rely on your rusty trig identities to invoke

2sin(θ)cos(θ) = sin(2θ)

 

[LowlyPion]

For the second problem the frictional force is just w(0.2) and acts in the direction resisting motion. The force that acts on the top rim acts at a distance to a pivot (I think) Possibly at the center of the glass.

Consider what the condition is for tipping?

Where will the center of mass necessarily need to be for it to fall?

 

[donotremember]

Consider what the condition is for tipping?

Where will the center of mass necessarily need to be for it to fall?

The center of mass must go beyond the vertical from the edge of the cup that remains on the table, but I'm unsure of how to include gravity and the two forces in order to understand how the system will evolve when you push on the cup.

This is a torque question isn't it?

 

[LowlyPion]

The center of mass must go beyond the vertical from the edge of the cup that remains on the table, but I'm unsure of how to include gravity and the two forces in order to understand how the system will evolve when you push on the cup.

This is a torque question isn't it?

Yes that's right. It's a question of torque.

You have the geometry of the glass. You have the maximum force that can be applied to the top without it slipping from the friction relationship.

If the diameter of the glass is d, then the center of mass is acting through what point due to gravity? And that gives you the moment from the center of mass to the leading edge about which it will tip. Now the force is acting at the top and that is at height h, so you have the tipping moment.

So long as you have a greater tipping moment, than settling moment from gravity, then it should tip all the way right? Because once it starts to lift, the center of mass acts through a shorter distance to the tipping edge and the force acts at a lengthening distance (height) from that same pivoting edge.