Q: Maximum weight for a suction cup hook on a glass door?

[tucky]

First off, I would like to thank HallsoIvy and Loren Booda for answering my last question. I really appreciate the help! However, I am really confused again. I have two problems that I need help with.

Q:
1. trailer measures 9 feet tall (ground-to-top) and 6 feet wide. It weighs 5 tons. The c.m. point is 4.5 ft. above the ground. What is the smallest radius curve this trailer can round at 60 mph without tipping?
A: convert to metric: mass=4539.26kg; v=26.82m/s; height=2.7432m; width=1.8288m; central mass=1.3716m

I was thinking about using the equation: Central Force=m*(v^2/r). However, I do not have the Central force….is there a way to get that or am I on the wrong track?

Q: A clear vinyl “suction cup” is stuck to a vertical glass door on a shower stall. The cup measures 1.5 inch in diameter. The coefficient of static friction between glass and clear vinyl is 0.4. The cup has a hook on it from which items may be hung. What is the maximum weight in lbs that can be hung from the cup if the cup is not to slide down the door?

A: A=pi r^2= pi(.77^2)=1.75in^2 14.7=psi air inside the suction cup
F=pA
F=14.7*.4*1.72in^2=10.11lb

 

[reinhard_t]

I think the question is flaw since you never mention ,whether the trail moves in horizontal or vertical plane...and also it cannot be done both way.

(so I assume probably you ask in horizontal case)

in horizontal case ,it needs resistance to tip

in the second question,I don't get the picture of the question.sorry about it.

 

[Doc Al]

Originally posted by tucky
I was thinking about using the equation: Central Force=m*(v^2/r). However, I do not have the Central force….is there a way to get that or am I on the wrong track?

You are on the right track, but it's tricky. Consider the forces acting on the truck: On the tires you have friction acting inward, and the normal force acting upward; plus you have the weight of the truck acting down. (I assume the road is horizontal and flat--not banked.)

Note that the normal force exerted on the inside tires goes to zero as the truck begins to tip: that's the condition for tipping. The easy way to solve this it to use the frame of reference of the truck, which is non-inertial: you must include a fictious force acting outward. The fictious force is the "centrifugal" force = mv²/r. Now apply the conditions for rotational equilibrium. Using the outer tires as the axis of rotation: The torque of the weight is balanced by the torque of the fictious force at the point when the truck just starts to tip.

You did fine on the second problem.

 

[reinhard_t]

first of all I will apologise my previous statement since I was very careless and just gave a glance...

now I have worked out together with the solution which you can check at the webpage below

http://www.angelfire.com/ultra/reinhard/tucky.html

!if the answer writing is not clear(since I use bad scanner),you can ask me to what I write...

sorry for the second question.I cannot do,because I don't know what the question talking...

hope the answer can benefit you.

 

[Doc Al]

Excellent reinhard_t. We get the same answer.

 

[tucky]

Thank you both! I really appreciate the help.