Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
Explanation:
From the question we are told that
Mass of ball 
Length of string 
Wind force 
Generally the equation for 
is mathematically given as
Max angle =
Generally the equation for max Height 
is mathematically given as
Generally the equation for Equilibrium Height 
is mathematically given as
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
- the net horizontal force acting on the beam is
where 
are the magnitudes of the tensions in ropes 1 and 2, respectively;
- the net vertical force acting on the beam is
where 
and 
.
Eliminating 
, we have
Solve for .
The answer is control variable
In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction.
AMA (Actual mechanical advantage) is found by dividing output force by effort force. The actual mechanical advantage will always be less than the ideal mechanical advantage. The ideal mechanical advantage assumes perfect efficiency which doesn't account for friction, while actual mechanical advantage does. Therefore; the IMA is always greater than the actual mechanical advantage because all machines must overcome friction.
Answer:
a) 0.022%
b) 10014.32 lb
Explanation:
a) Percentage uncertainty would be
Percent uncertainty is 0.022%
b) For 1 kg uncertainty mass in kg would be
Mass in pounds would be
Mass in pound-mass is 10014.32 lb
