PHYSICS I. CHAPTER 12 AND 13. 1. An 82.0 kg-diver stands at the edge of a light
ID: 1601836 • Letter: P
Question
PHYSICS I. CHAPTER 12 AND 13.
1. An 82.0 kg-diver stands at the edge of a light 5.00-m diving board, which is supported by two pillars 1.60 m apart.
a) Find the force exerted by pillar A.
b) Find the force exerted by Pillar B.
QUESTION 1 COMES WITH A DIAGRAM. I JUST COULDN'T PASTE IT HERE. YOU WOULD BE ABLE TO FIND IT BY COPYING QUESTION 1 AND PASTING IT ON CHEGG. YOU SHOULD SEE AN OPTION OF THE QUESTION THAT COMES WITH THE DIAGRAM.
2. A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm mark. What is the mass of the meter stick?
3. A shear force of 400N is applied to one face of an aluminium cube with sides of 30 cm. What is the resulting relative displacement? (The shear modulus for aluminium is 2.5 * 10^(10) N/m^(2)
4. The tensile strength for a certain steel wire is 3000MN/m^(2). What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this kind of steel?
5. A satellite of mass 500 kg orbits the Earth with a period of 6000s. The Earth has a mass of 5.98 * 10^(24) kg.
a) Calculate the magnitude of the Earth's gravitational force on the satellite.
b) Determine the altitude of the satellite above the Earth's surface.
6. Three identical 50-kg masses are held at the corners of an equilateral triangle, 30 cm on each side. If one of the masses is released, what is its initial acceleration, if the only forces acting on it are the gravitational forces due to the other two masses?
NEED ANSWERS ASAP PLEASE!!! I WOULD ALSO APPRECIATE DETAILS ON HOW EACH PROBLEM WAS SOLVED.
Explanation / Answer
(1) Consider the forces of which there are three. the diver and the two supports. These must sum to 0.
Fa = force on pillar A
Fb = force on pillar B
Force of diver = mass*g = 82*9.8 = 803.6 N
Fa + Fb - 803.6 = 0
Pillar B produces an upward or clockwise torque = Fb*1.6
Diver produces a counterclockwise torque = 803.6*5 = 4018 N-m
The combined torques must be 0.
Fb*1.6 - 4018 = 0
Fb = 2511.25 N (upward)
Now for Fa we use the force equation:
Fa + Fb - 803.6 = 0
Fa = -1.71 kN
So the forces are:
Pillar A force = -1707.65 = 1.71 kN in a downward direction
Pillar B force = 2511.25 = 2.51 kN in an upward direction
(2) Since meter stick balances at {x = 61.3 cm} when
mass of {m = 50 g} is placed at {x = 90 cm}, the
torques from meter stick mass "M" and smallmass
{m = 50 g} must balance about {x = 61.3 cm}
(M*(61.3 - 0)/(100)) * ((61.3 - 0)/2) = (M*(100 - 61.3)/(100)) * ((100 - 61.3)/2) + (50)*(90 - 61.3)
Solving for "M",
Meter Stick Mass = M = 127 grams
(3) From the shear deformation and the shear modulus is
F = S*(del_X/Lo)*A
Area of the cube (A) =Lo^2
Now
F = S*(del_X)*Lo ....................(1)
Given that
A shear force of (F) = 400 N
Length of the side (Lo) = 30 cm = 0.3 m
The shear modulus for aluminum is (S) = 2.5*10^10 N/m^2
The from the equation (1) we get the resultant relative displacement is
del_X = F/S*L0
= 400/2.5*10^10* 0.3 = 5.33*10^-8 m
(4) Max load = 3000*10^6 *pi* (3*10^-3)^2 / 4
= 21.205 kN
Therefore, the answer is 21.205 kN
(5) Mass of the earth Me = 5.98*10^24 kg
mass of the satellite ms = 500 kg
Time period T = 6000 sec
Gravitational constant G = 6.67*10^-11 N m^2/kg^2
Radius of the earth Re = 6.38*10^6 m
(a) calculate the distance between center of earth and satellite.
T^2 = 4*pi^2*r^3/G*Me
6000^2 = 4*pi^2*r^3/6.67*10^-11*5.98*10^24
r = 7.14*10^6 m
calculate the earth gravitational force
F = G*Me*ms/r^2
= 6.67*10^-11*5.98*10^24*500/(7.14*10^6)^2
F = 3912.01 N
(b) calculate the altitude of the satellite above the earth surface
a = r - Re
= 7.14*10^6 - 6.38*10^6
= 760 km
(6) The equation we got to use for gravitational force due to each of the other masses is
F = G*m1*m2/r^2
= (6.67*10^-11)*50*50/0.30^2
= 1.85*10^-6 N
However since we have the force vectors at 60 degress to each other, the x-components of them will cancel
and the y-components will stay.
Fy = (1.85*10^-6)*Cos(30) = 1.60*10^-6 N
However since there are two forces we got to add their y-components together, which gives us 3.20*10^-6 N.
To find acceleration, we simply use F = m*a and solve for a.
a = F/m =(3.20*10^-6)/50
= 6.41*10^-8 m/s^2
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