# BasilMarket

## Chat Forum Talk about topics not related to MapleStory

Level 180 Scania Zero Transcendent
I need help with the following problems, I forget how to do them.

What horizontal force must be applied to a large block of mass M shown in the figure below so that the tan blocks remain stationary relative to M? Assume all surfaces and the pulley are frictionless. Notice that the force exerted by the string accelerates m2. (Use the following as necessary: m1, m2, M, and g.)

The mass of a roller-coaster car, including its passengers, is 486 kg. Its speed at the bottom of the track in the figure below is 21 m/s. The radius of this section of the track is r1 = 24 m. Find the force that a seat in the roller-coaster car exerts on a 48-kg passenger at the lowest point.

A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the horizontal total force on the driver has magnitude 141 N. What is the total horizontal force on the driver if the speed on the same curve is 21.0 m/s instead?

A coin placed 29.8 cm from the center of a rotating, horizontal turntable slips when its speed is 48.8 cm/s.What is the coefficient of static friction between coin and turntable?

Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly hump" and had it installed. Suppose a 1 800-kg car passes over a hump in a roadway that follows the arc of a circle of radius 21.4 m as in the figure below.If the car travels at 29.5 km/h what force does the road exert on the car as the car passes the highest point of the hump?What is the maximum speed the car can have without losing contact with the road as it passes this highest point?

# Replies

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Level 180 Scania Zero Transcendent
The mass of a roller-coaster car, including its passengers, is 486 kg. Its speed at the bottom of the track in the figure below is 21 m/s. The radius of this section of the track is r1 = 24 m. Find the force that a seat in the roller-coaster car exerts on a 48-kg passenger at the lowest point.

Isn't this just F=ma?
F=(9.8)(48)[/quote]

I tried that already and it says my answer is close. Said like it's within 10% range of the correct answer or something like that. This is an online assignment.
Feb 20 2012
Level 184 Khaini Corsair
Send me the picture of the figures and I'll do the problems for you. If you can't get me the figures then I'm not bothering trying to imagine it.
Feb 20 2012
Next time don't wait until the last minute.
Feb 20 2012
I tried that already and it says my answer is close. Said like it's within 10% range of the correct answer or something like that. This is an online assignment.[/quote]

I don't know for sure then Perhaps the online assignment used a=10 m/s2 instead of 9.8?

A coin placed 29.8 cm from the center of a rotating, horizontal turntable slips when its speed is 48.8 cm/s.What is the coefficient of static friction between coin and turntable?
There is an equation for this, but I forgot it. It depends on Fnormal though.
I dunno what the first question diagram looks like and i have no idea how speed bumps work
Feb 20 2012
idk but i found this on yahoo answers with similar number

A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the horizontal total force on the driver has magnitude 129 N. What is the total horizontal force on the driver if the speed on the same curve is 17.1 m/s instead?

F is proportinal to v^2
F2/F1 = (v2/v1)^2 = (17.1/14)^2
F2 = 129*(17.1/14)^2 = 405 N

good luck
Feb 20 2012
Level 71 Khaini Mechanic 3
3mins!
Feb 20 2012
11 minutes [/quote]
2 min.
Feb 20 2012
Level 71 Khaini Mechanic 3
1min!
Feb 20 2012
Level 222 Bera Hero
I tried that already and it says my answer is close. Said like it's within 10% range of the correct answer or something like that. This is an online assignment.[/quote]

Also don't forget that centripetal force is proportional to v^2/r.
Feb 20 2012
Level 184 Khaini Corsair
Roller Coaster problem:

They gave you the radius therefore we're using: (0.5mv^2)/r = mgh