Calculating Mary's Moped Speed: Solving for Maximum Velocity

  • Thread starter plane
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In summary, the question asks for the maximum speed of Mary's moped as she accelerates to maximum speed in 18 seconds. Mary dangles her watch from a thin piece of string, making an angle of 15 degrees with respect to the vertical. The only forces acting on the watch are the force of gravity and the tension force, Ft. By resolving Ft into horizontal and vertical components, and equating the vertical component to the gravitational force, an expression for Ft is obtained. Using this, the acceleration can be calculated by solving the equation Ftx= m(ax), resulting in a maximum speed of gtan15.
  • #1
plane
19
0
hey guys i am stumped on this question:

mary dangles her watch from a thin piece of string, while burning rubber on her supercharged moped. she notices that the string makes an angle of 15degrees with respect to the vertical while the moped accelerates to maximum speed, which takes about 18s. what is the maximum speed of mary's moped?

i am really stuck on this question and i need to show my work too. can anyone please help? thanks.
 
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  • #2
What, exactly, have you done so far?
 
  • #3
All you have to do is work out the acceleration on the watch/moped.
The watch is at an angle of 15 deg.
Draw a diagram of the forces on the watch. What are the only forces acting ?
From this, you should be able to get the acceleration.
 
  • #4
i can't get anything, i just keep staring at the diagram but i can't figure it out, how do i figure out the acceleration by using the angle? do i brake it into the X and Y components? then what? :cry:
 
  • #5
is there no one that can help me?
 
  • #6
Fermat said:
All you have to do is work out the acceleration on the watch/moped.
The watch is at an angle of 15 deg.
Draw a diagram of the forces on the watch. What are the only forces acting ?
From this, you should be able to get the acceleration.

the only forces acting are the force of gravity (mg) and Ft (tension)
 
  • #7
Have you got the diagram ?

What forces have you got labelled on it ?
 
  • #8
Ah, already answered.

So, now resolve the Tension, Ft, into horizontal and vertical components.
 
  • #9
Fermat said:
Ah, already answered.

So, now resolve the Tension, Ft, into horizontal and vertical components.


how? it only gives me the angle and the time, no forces or mass of the watch.
 
  • #10
If you know the angle the tension is acting at, can't you just then resolve the tension, call it Ft, into horizontal and vertical components using that angle ?


Take the mass of the watch as m.
 
Last edited:
  • #11
Fermat said:
If you know the angle the tension is acting at, can't you just then resolve the tension, call it Ft, into horizontal and vertical components using that angle ?

Where did the time come from ?

Take the mass of the watch as m.

so this is what id be doing:

Ftcos15= Ftx
Ftsin15= Fty

and then Ftx= m(ax)?

then what do i substitute?

the time comes from:
accelerates to maximum speed, which takes about 18s. what is the maximum speed of mary's moped?

the question gives it to me
 
Last edited:
  • #12
Almost there.

Ft.cos15 is actually the vertical component.

You can equate this to the only other vertical force, the mass of the watch, hence you can get an expression for Ft.
 
  • #13
Fermat said:
Almost there.

Ft.cos15 is actually the vertical component.

You can equate this to the only other vertical force, the mass of the watch, hence you can get an expression for Ft.

i always thought sine was the vertical componenet, when resolving vectors and i think that's true. also it doesn't give me the mass so how can i get an expression for Ft? can you please show me how?
 
  • #14
I'll upload a sketch.

The masss, m, should cancel out.,
 
  • #15
See attachment.
 

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  • #16
What is your expression for Ft ?
 
  • #17
Fermat said:
What is your expression for Ft ?

your attachment is pending approval, as for Ft i have no idea how to get an expression for it.
 
  • #18
You had two eqns,

Ftcos15= Ftx
Ftsin15= Fty

but they should have been,

Ftsin15= Ftx
Ftcos15= Fty

(You will see why the cos and sin are the other way around when the attachment is approved. But it's just really because the angle, alpha, is the angle to the vertical. If it had been the aqngle to the horizontal, then sin would have been the vertical component.)

The only vertical force is the mass of the watch, so Fty = mg. This gives,

Ftcos15= Fty = mg
==============

Now you have an expression for Ft, can you solve Ftx= m(ax) for a ?
 
  • #19
Fermat said:
You had two eqns,

Ftcos15= Ftx
Ftsin15= Fty

but they should have been,

Ftsin15= Ftx
Ftcos15= Fty

(You will see why the cos and sin are the other way around when the attachment is approved. But it's just really because the angle, alpha, is the angle to the vertical. If it had been the aqngle to the horizontal, then sin would have been the vertical component.)

The only vertical force is the mass of the watch, so Fty = mg. This gives,

Ftcos15= Fty = mg
==============

Now you have an expression for Ft, can you solve Ftx= m(ax) for a ?

you have written an expression for FTy, but i need Ftx, is Ftx the same formula that being:
Ftsin15=Ftx=m(ax)?
 
  • #20
[tex]F_tsin15 = F_{t_x} = m(a_x)[/tex]
[tex]F_tcos15= F_{t_y} = mg[/tex]

[tex]F_tsin15 = m(a_x)[/tex]
[tex]F_tcos15 = mg[/tex]

Divide 1st eqn by 2nd eqn,

[tex]\frac{sin15}{cos15} = \frac{a_x}{g}[/tex]

[tex]a_x = gtan15[/tex]
 
  • #21
Fermat said:
[tex]F_tsin15 = F_{t_x} = m(a_x)[/tex]
[tex]F_tcos15= F_{t_y} = mg[/tex]

[tex]F_tsin15 = m(a_x)[/tex]
[tex]F_tcos15 = mg[/tex]

Divide 1st eqn by 2nd eqn,

[tex]\frac{sin15}{cos15} = \frac{a_x}{g}[/tex]

[tex]a_x = gtan15[/tex]

thanks i get it now
 

1. What is the formula for calculating maximum velocity?

The formula for calculating maximum velocity is Vmax = √(Fmax/m), where Vmax is the maximum velocity, Fmax is the maximum force, and m is the mass of the object.

2. How do I determine the maximum force in the equation?

The maximum force can be determined by using the equation Fmax = ma, where Fmax is the maximum force, m is the mass of the object, and a is the acceleration.

3. What units should be used for the variables in the formula?

The units for the maximum velocity will depend on the units used for the maximum force and mass. Common units for maximum velocity include meters per second (m/s) or kilometers per hour (km/h).

4. Can this formula be used for any object's maximum velocity?

Yes, this formula can be used for calculating the maximum velocity of any object as long as the maximum force and mass are known.

5. Are there any other factors that could affect the maximum velocity of an object?

Other factors that could affect the maximum velocity of an object include air resistance, friction, and the object's shape and size. These factors may need to be taken into consideration for a more accurate calculation.

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