A man of mass M kg and his son of mass m kg are standing in a lift. When the lift is accelerating upwards with magnitude 1ms-2 the magnitude of the normal contact force exerted on the man by the lift floor is 880N. When the lift is moving with constant speed the combined magnitude of the normal contact forces exerted on the man and the boy by the lift floor is 1000N. Find the values of M and m.
forceonman=880=Mg+ma=M(9.8+1)
forceonmanconstantspeed=1000=(M+m)g
well, from the Man data, M = 880/(10.8)
put that into the second equation, and solve for m, the son mass.
AS 880N IS THE FORCE EXERTED BY THE LIFT TO THE MAN
THEREFORE,880-Mg=Ma
880=M(g+a)
880/g+a=M
880/11=80=M
M=80kg
880+N2=1000
N2=120N
N2=m(g+a)
120/11=m
m=11kg
Let's break down the problem step-by-step:
Step 1: Analyze when the lift is accelerating upward -
When the lift is accelerating upward, the total force acting on the man and his son can be calculated using Newton's second law:
Total force = (mass of man + mass of son) × acceleration
The weight of the man and his son forms a downward force given by:
Weight = (mass of man + mass of son) × g
where g is the acceleration due to gravity.
To find the normal contact force exerted on the man by the lift floor, we subtract the weight of the man and his son from the total force:
Normal force on man = Total force - Weight
Given that the normal force on the man is 880N, we can write the equation as:
880N = (mass of man + mass of son) × 1ms⁻² - (mass of man + mass of son) × g
Step 2: Analyze when the lift is moving at a constant speed -
When the lift is moving at a constant speed, the net force acting on the man and his son is zero. The weight of the man and his son is balanced by the normal contact force exerted on them by the lift floor.
Given that the combined magnitude of the normal contact forces is 1000N, we can write the equation as:
1000N = (mass of man + mass of son) × g
Step 3: Solve the system of equations -
We have two equations:
1) 880N = (mass of man + mass of son) × 1ms⁻² - (mass of man + mass of son) × g
2) 1000N = (mass of man + mass of son) × g
To solve for M and m, we can use substitution. Rearranging equation 2, we get:
(g) = 1000N / (mass of man + mass of son)
Substituting this value of g into equation 1, we get:
880N = (mass of man + mass of son) × 1ms⁻² - (mass of man + mass of son) × (1000N / (mass of man + mass of son))
Simplifying further:
880N = (mass of man + mass of son) × (1ms⁻² - 1000N / (mass of man + mass of son))
Let's call (mass of man + mass of son) as M_total.
880N = M_total × (1ms⁻² - 1000N / M_total)
Multiply both sides by M_total:
880N × M_total = M_total × (1ms⁻² - 1000N / M_total) × M_total
880N × M_total = 1ms⁻²M_total - 1000N
Rearranging:
880N × M_total + 1000N = 1ms⁻²M_total
Combining like terms:
(880N + 1000N) = 1ms⁻²M_total
1880N = M_total × 1ms⁻²
Divide both sides by 1ms⁻²:
M_total = 1880N / 1ms⁻²
So, the total mass (M_total) is equal to 1880 kg.
Now, substituting the value of M_total into equation 2:
1000N = M_total × g
1000N = 1880 kg × g
Divide both sides by 1880 kg:
g = 1000N / 1880 kg
g ≈ 0.532 ms⁻²
Now we can substitute the value of g back into the equation for M_total:
M_total = 1880N / 0.532ms⁻²
M_total ≈ 3533 kg
Finally, we can substitute the value of M_total back into the equation for the individual masses:
M + m = M_total
M + m ≈ 3533 kg
Since we don't have any other information about M and m, we cannot determine their exact values. However, we know that the sum of their masses is approximately 3533 kg.
To solve this problem, we need to use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's analyze each situation separately to find the values of M and m.
1. When the lift is accelerating upwards, the net force on the man is the gravitational force acting downwards and the normal force exerted by the floor of the lift upwards. We can write this as:
Net force = Mg - N = Ma
Where Mg is the weight of the man (M * g), N is the normal force exerted by the floor, a is the acceleration of the lift, and Ma is the net force acting on the man.
Given that the acceleration of the lift is 1 m/s^2 and the normal force N is 880 N, we can solve for M:
880 = M * 9.8 - M * 1
880 = 9.8M - M
880 = 8.8M
M = 880 / 8.8
M = 100 kg
So, the mass of the man is 100 kg.
2. When the lift is moving with constant speed, there is no acceleration, so the net force on the man and the boy is zero. This means that the gravitational force and the normal forces are balanced:
Net force = Mg - N = 0
Given that the combined magnitude of the normal contact forces (N) is 1000 N, we can solve for the total mass (M + m):
1000 = (M + m) * 9.8
Since we already know the value of M from the previous calculation, we can substitute it in:
1000 = (100 + m) * 9.8
Solving for m:
1000 = 980 + 9.8m
9.8m = 1000 - 980
9.8m = 20
m = 20 / 9.8
m ≈ 2.04 kg
So, the mass of the son is approximately 2.04 kg.
Therefore, the values of M and m are M = 100 kg and m ≈ 2.04 kg, respectively.