Physics help. Please respond?

Question

How do I work this problem?

A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.28. Suddenly, the plane accelerates, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

Can you please help me with this?

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I've asked this question earlier and still haven't gotten an answer. I've been told how to do it but it's still confusing me b/c I have no idea what the ans is.

Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 3.60 m. Two of the spheres have a mass of 4.50 kg each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?

Answer 1

Hi lovinit

No disrespect to the professor, but both questions can be answered as asked because objects fall at the same rate in gravitaitonal fields regardless of their mass.

Problem 1:
Four forces are acting on the cup at the moment the plane accelerates. Draw a picture of the cup and draw four arrows, one each pointing up, down, forwards and backwards. The forces are:
* the force acting downwards we'll call the normal force, N. This is supplied by gravity: N = mg (for m the mass of the cup).
* the force acting upwards is supplied by the table and exactly balances the normal force - the cup doesn't move up or down
* the force acting backwards is the friction force F. This force is the product of the coefficient of friciton and the normal force: F = uN
* the force acting forwards is supplied by the acceleration of the plane A. This force is equal to the product of the cup's mass and the acceleration on it: A = ma

The question asks us to balance A and F for maximum acceleration (a). So:
A = F
ma = uN
ma = umg
a = ug

The maximum acceleration the plane can provide to the cup before it tips is equal to the coefficient of friction (u) multiplied by the acceleration due to gravity (g).
a = 0.28*9.8
a = 2.7m/s/s


Problem 2:
To attack this problem we work out the acceleration of the third sphere towards each of the other spheres, and then resolve the resultant acceleration vector. The magnitude of the acceleration of sphere three (m) to each other sphere (M) is given by:

gravitational force = accelerating force
GMm/r^2 = ma
GM/r^2 = a

You can substitute the values for G (gravitational constant), M (the mass of sphere one or two) and r (the separation between sphere 3 and sphere 1 or 2) from your question to obtain a.

Now you have two vectors of equal magnitude pointing from sphere 3, one towards sphere 1 and the other at sphere 2. The angular separation of the two vectors is 60 degrees. You need to add the vectors to obtain the resultant a vector, which will point between spheres 1 and 2. I presume you know how to add vectors - just translate one until the two vectors are top to tail, and compute the resultant length: = 2*a*cos30 = a*sqrt(3) (where a is the answer you obtained above).


Hope this helps!
The Chicken

Answer 2

1. f=ma. now the cup is trying to mve towards you with a force ma where m is the mass of the cup and a the accelration of the plane. now the cup is acted upon by g the gravitational force.mg.
the force of friction acting in the opposite direction to the movement of the cup is 0.28mg. so ma=0.28mg or a = 0.28mg/m=0.28g.

2. F=Gm1m2/d^2. you need a figure for this. the gravitational force acting due to each individual sphere is G m*4.5/3.6^2 cos30
to get for both you simply multiply by two.

Answer 3

First, you need to know the mass mass of the coffee cup since mu (.28) = F / N

For the spheres, you also need the mass of the third sphere since F = [G (m1 * m2) / d^2]...d = sqrt (3.6)m

and: F=ma