Answer: The statement first and the fourth statement are true.
Explanation:
According to Newton's gravitational law, every particle in the universe attracts every other particle with the force of attraction between the masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
As we move to higher altitude, the force of gravity on use decreases because the force of gravity is inversely proportional to the distance.
If the masses of the two objects are more then there will be greater force of gravity between them.
Therefore, the statement first and the fourth statement are true.
Answer:
A) The new amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please find the attached files for the solution
This problem has three questions I believe:
>
How hard does the floor push on the crate?
<span>We have to find the net
vertical (normal) Fn force which results from Fp and Fg.
We know that the normal component of Fg is just Fg, which is equal to as 1110N.
From the geometry, the normal component of Fp can be calculated:
Fpn = Fp * cos(θp)
= 1016.31 N * cos(53)
= 611.63 N
The total normal force Fn then is:
Fn = Fg + Fpn
= 1110 + 611.63
=
1721.63 N</span>
> Find the friction
force on the crate
<span>We
have to look for the net horizontal force Fh which results from Fp and Fg.
Since Fg is a normal force entirely, so we can say that the
horizontal component is zero:
Fh = Fph + Fgh
= (Fp * sin(θp)) + 0
= 1016.31 N * sin(53)
=
811.66 N</span>
> What is the minimum
coefficient of static friction needed to prevent the crate from slipping on the
floor?
We just need to compute the
ratio Fh to Fn to get the minimum μs.
μs = Fh / Fn
= 811.66 N / 1721.63 N
<span>=
0.47</span>
Answer:
Explanation:
An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, 
the angle of the slope
is the frictional force, with 
being the coefficient of friction and R the normal reaction of the incline
The equation of the forces along the direction perpendicular to the slope is
where
R is the normal reaction
is the component of the weight perpendicular to the slope
Solving for R,
And substituting into (1)
Re-arranging the equation,
This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.
