a 60kg man is on a steadily rotating Ferris wheel. the man is standing on a bathroom scale and the scale reads 45kg at the top of the wheel. take g=10m/s^2
(a) what does the scale read when the man is at the bottom of the wheel?
(b)what does the scale read when the man is level with the hub of the wheel going up? assume their is sufficient friction that the man's feet do not slide off the scale?
(c) what co-efficient of friction is needed to keep the man's feet on the scale in (b)?
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i was able to figure out the acceleration which is -2.5 m/s^2..for part A the answer is 75kg however i don't understand how to get that answer as well as the other two parts...any help would be appreciated :)
b) the man is vertical, his weight is mg, centripetal force does not add to weight. However, to keep him accelerating inward, fricion must be m*v^2/r
To solve this problem, we will use the concept of centripetal force and the gravitational force acting on the man.
(a) When the man is at the bottom of the Ferris wheel, the gravitational force and the centripetal force act in the same direction. The net force will be the sum of these two forces.
Let's consider the forces acting on the man at the bottom of the wheel:
- Gravitational force (Fg) = mg (where m is the mass of the man and g is the acceleration due to gravity)
- Centripetal force (Fc) = m * a (where a is the acceleration)
Since the net force is the sum of these forces, at the bottom of the wheel, we have:
Net force = Fg + Fc = mg + ma
Using the value of m = 60 kg and g = 10 m/s^2, we can calculate the net force:
Net force = (60 kg) * (10 m/s^2) + (60 kg) * (-2.5 m/s^2) = 600 N - 150 N = 450 N
The scale reading will be equal to the net force, so the scale will read 45 kg + (450 N / 10 m/s^2) = 45 kg + 45 kg = 90 kg.
Therefore, the scale will read 90 kg when the man is at the bottom of the wheel.
(b) When the man is level with the hub of the wheel going up, the gravitational force and the centripetal force are in opposite directions. The net force will be the difference between these two forces.
Again, let's consider the forces acting on the man at this position:
- Gravitational force (Fg) = mg
- Centripetal force (Fc) = m * a
The net force would be:
Net force = Fg - Fc = mg - ma
Using the same values as before, the net force is:
Net force = (60 kg) * (10 m/s^2) - (60 kg) * (-2.5 m/s^2) = 600 N + 150 N = 750 N
The scale reading will be equal to the net force, so the scale will read 45 kg + (750 N / 10 m/s^2) = 45 kg + 75 kg = 120 kg.
Therefore, the scale will read 120 kg when the man is level with the hub of the wheel, going up.
(c) In order for the man's feet to stay on the scale, there must be enough friction force between the man's feet and the scale to counteract the gravitational force and prevent sliding off.
The maximum friction force can be given by Fmax = µ * N
where µ is the coefficient of friction and N is the normal force, which is equal to the gravitational force acting on the man.
Fmax = µ * mg
To keep the man's feet on the scale, the friction force should be greater than or equal to the centripetal force acting on the man when he is level with the hub going up. So we have:
Fmax ≥ Fc = ma
Substituting Fmax = µ * mg and Fc = ma, we get:
µ * mg ≥ ma
Dividing by m on both sides:
µ ≥ a/g
Plugging in values, we have:
µ ≥ (-2.5 m/s^2) / (10 m/s^2) = -0.25
Since the coefficient of friction cannot be negative, we take the positive value:
µ ≥ 0.25
Therefore, a coefficient of friction of at least 0.25 is needed to keep the man's feet on the scale.
To answer these questions, we'll use the concepts of centripetal force, gravitational force, and the normal force acting on the man. We can break the problem into different scenarios and analyze the forces acting in each case.
(a) When the man is at the top of the Ferris wheel, the centripetal force directed towards the center of the wheel is provided by the net force due to the gravitational force and the normal force. The normal force is acting upward, and we can write the equation:
Net force = Centripetal force
(mg + N) = m_a
where m = mass of the man, g = acceleration due to gravity, N = normal force, m_a = mass times acceleration (centripetal acceleration).
Given that the man's mass (m) is 60 kg, the acceleration due to gravity (g) is 10 m/s^2, and the reading on the scale (N) is 45 kg, we can substitute these values into the equation above to solve for the centripetal acceleration (m_a):
(60 kg * 10 m/s^2) + 45 kg = m_a
600 kg·m/s^2 + 45 kg = m_a
645 kg·m/s^2 = m_a
Since the Ferris wheel is rotating steadily, the centripetal acceleration (m_a) is constant. Therefore, the equation still holds true when the man is at the bottom of the wheel.
(mg + N') = m_a
However, at the bottom of the wheel, the normal force (N') and the gravitational force are acting in opposite directions, which affects the sign of the net force. As the man is at the bottom, the net force is given by:
Net force = Centripetal force
(N' - mg) = m_a
To find the reading on the scale (N'), we need to solve for it using the equation above. We can rearrange the equation:
N' = m_a + mg
Now we can substitute the known values:
N' = 645 kg·m/s^2 + (60 kg * 10 m/s^2)
N' = 645 kg·m/s^2 + 600 kg·m/s^2
N' = 1245 kg·m/s^2
Therefore, the scale reading at the bottom of the wheel is 1245 kg.
(b) When the man is level with the hub of the wheel and going up, there are two forces acting: the gravitational force (mg) and the normal force (N''). These forces together provide the centripetal force:
Net force = Centripetal force
(N'' + mg) = m_a
Since the man is at the same level as the hub, there is no change in the net force due to the gravitational force; therefore:
N'' = m_a
Substituting the known value of the centripetal acceleration (m_a = 645 kg·m/s^2):
N'' = 645 kg·m/s^2
Therefore, the scale reading at the level with the hub going up is 645 kg.
(c) To find the coefficient of friction needed to keep the man's feet on the scale when at the level of the hub going up, we need to consider the friction force acting opposite to the direction of motion.
Friction force (f) = μN''
where μ is the coefficient of friction, and N'' is the normal force at the level with the hub going up. We can take the mass (m) of the man as 60 kg and the acceleration due to gravity (g) as 10 m/s^2.
The normal force at the level with the hub going up (N'') is equal to the gravitational force:
N'' = mg
Substituting these values, we get:
f = μmg
Since the man is not sliding off the scale, the maximum frictional force must be equal to the gravitational force (mg). Thus, the equation becomes:
μmg = mg
We can cancel out the mass and acceleration due to gravity on both sides of the equation:
μ = 1
Therefore, the coefficient of friction needed to keep the man's feet on the scale is 1.