Rod's Physics 3/4 Questions Thread

[Rod]

Hey guys just need some help on this question;

A car of mass 1280 kg travels around a bend with a radius of 12m. The total sideways friction on the wheel is 16400N. Calculate the maximum constant speed of the car if the road is banked at an angle of 10 degrees to the horizontal.


As in maximum speed, I assume it means the highest speed it can travel before skidding off the banked curve. So two key formulas I would use when doing this question;

Fnet=mv2/r

Fnet=Frcos0+ Nsin0 (0=theta I don't know how to do theta on AN)

Values we have:
m=1280 kg, r=12m, friction=16400n, normal = ? fnet = ? v=?

Fnet=Frcos0+ Nsin0
Fnet= 16400cos10 + Nsin10

N= Wcos0
N = 10x1280xCos10
N= 12606

So fnet= 16400cos10 + 12606sin10
Fnet= 18339 N

So Fnet=mv2/r
So 18339= 1280v2/12
18339x12/1280 = v2
v2= 172
v= 172 (square rooted)
v- 13m/s-1

What am I doing wrong here?

Thanks again guys, much appreciated

[Rod]

The answer in our holiday homework booklet is 26.4 m/s-1

[SocialRhubarb]

The error you've made is that you assumed .

The normal force of the road on the car can change as a result of a variety of factors on the car, which is why it is often referred to as a normal 'reaction' force. The normal is a reaction to the forces applied onto the plane by objects.

In this case, because the car is accelerating towards the centre of the bend in the road, the normal force increases, and hence we cannot assume that , since this only holds when the mass is stationary, or accelerating parallel to the plane.

In order to answer this question, we resolve the forces into horizontal and vertical components. You've already done the horizontal component, so that shouldn't be a problem.

The car is not moving upwards or downwards, so we can say that its vertical acceleration is 0, and hence the net force in the vertical direction is 0, so the sum of all the forces acting in a vertical direction must be 0. If we equate the vertical components of the normal force, frictional force and weight force, and equate them to 0, we should get a different value for the normal force, and substituting this back into your original expression, you should have a different answer.

[Rod]

Ok so I not only have been having trouble with this question, but also I have some few questions on the theory in regard to the following question.

A 50 kg circus performer grips a vertical rope with her teeth and sets herself moving to a circle with a radius of 5 m at a constant horizontal speed of 3m/s.

a) What angle does the rope make the vertical?

Okay, so let's figure out the fnet first and then use it to find the angle.. Fnet= mv2/r, so 50x30^2/5 = 90 N. Now, if we imagine this guy biting a rope and swinging in circles, we can make a triangle, in where the hypotenuse in the tension (tension of the rope), the adjacent is the weight while the opposite side is the fnet. If so we can go, fnet/w=tantheta, and then use fnet, tan and the weight, and eventually figure out the angle to be 10 degrees. This is wrong, it's supposed to be 5 degrees, what am I doing wrong here?

In the next bit you need to find the tension. And by using the triangle we pictured I ended up correctly finding the value of tension in the rope. So why isn't the angle correct?

Thanks guys

[lzxnl]

I can't see anything wrong with what you've done.

[Rod]

I can't see anything wrong with what you've done.

The answer is 4.8 degrees. I re-calculated it like five times. You reckon the suggested answer is wrong?

Thanks

[hongkyho]

eventually figure out the angle to be 10 degrees

I got around 10 degrees as well.

In the next bit you need to find the tension. And by using the triangle we pictured I ended up correctly finding the value of tension in the rope.

Based on this, I'm pretty sure the answers are wrong as well.

[Rod]

Hey guys,

I just can't seem to get my head around analog and digital devices. So here is what I know; for analog devices, they can respond and process information over a range of voltages. Also, they represent their information by symbols (like an analog clock). For digital, they can only respond to two values, high or low voltages, and display they information by using numbers. Is this all I should know? I feel that I need to know more. Why can't digital devices respond to a range of voltages? What other features of digital make it so much superior to analog devices? What features of analog devices make it such an out dated device?

Thanks everyone

[Rod]

Hey guys sorry for disturbing again,

With output voltage. Say I figure out the output voltage to be 4V, does that mean I can use that voltage and the resistance of r2 to figure out the current of R2? Does that also mean, if the input voltage is 6v, the output voltage for r1 will be 6-4=2v? Just a bit confused.

Thanks again.

[PB]

^You are precisely right. as for the post above that...I am not sure I quite understand your question. I don't think analogs and digital etc. are on the course

[Rod]

^You are precisely right. as for the post above that...I am not sure I quite understand your question. I don't think analogs and digital etc. are on the course

Ohhh my gooddd I am going to BURN my text book haha! I should keep the study design next to me shouldn't I ?

Thanks Pb!

[lzxnl]

Ohhh my gooddd I am going to BURN my text book haha! I should keep the study design next to me shouldn't I ?

Thanks Pb!

If you EVER think something seems abnormally hard in the course, check the study design. 7 times out of 10, it won't be on the course. Just a rough estimate but you get the point

[PB]

haha, you are welcome Rod   The thing with the study design is that it can be ambiguous and overly-simplistic at times, and hence, may not be of much help in determining what is on the course.
I think a better method of verifying course material may be to look at the questions in trustworthy study guides like Neap/Checkpoints etc.
Hope that helps

[Rod]

Hey guys, another question here;

So a diode is placed in a circuit. I need to calculate the current flowing through the circuit in mA.

The circuit is very simple, it shows a battery that supplies 6 violts, a diode and a 200 ohm resistor, the circuit is in series. Next to the circuit is a characteristic curve of the diode, the curve shows that at 0.7 volts the curve starts increasing, meaning the voltage drop of the diode is 0.7 volts.

So, with that information I was able to calculate that the voltage going through the resistor is 6-0.7, 5.3 volts. The resistor is 200 ohm. So I=v/r, 5.3/200 = 0.027 amps.

Since the question wants it in milli amps I times it by 10^-3 to get 0.000027 mA.

ALTHOUGH, the answer is 26.5 amps??

Can someone please tell me what I am doing wrong here.

Thanks again for the help,

Rod

[lzxnl]

Hey guys, another question here;

So a diode is placed in a circuit. I need to calculate the current flowing through the circuit in mA.

The circuit is very simple, it shows a battery that supplies 6 violts, a diode and a 200 ohm resistor, the circuit is in series. Next to the circuit is a characteristic curve of the diode, the curve shows that at 0.7 volts the curve starts increasing, meaning the voltage drop of the diode is 0.7 volts.

So, with that information I was able to calculate that the voltage going through the resistor is 6-0.7, 5.3 volts. The resistor is 200 ohm. So I=v/r, 5.3/200 = 0.027 amps.

Since the question wants it in milli amps I times it by 10^-3 to get 0.000027 mA.

ALTHOUGH, the answer is 26.5 amps??

Can someone please tell me what I am doing wrong here.

Thanks again for the help,

Rod

OK your unit conversion is slightly iffy. 1 A = 1000 mA
Multiplying both sides by 0.027, we get 0.027 A = 27 mA
That's a bit closer to the answer you have. In fact, if you didn't round 5.3/200=0.0265, you would end up with the same answer.

Just remember that 1 A = 1000 mA and that you'll always have more mA than A.