A bicycle is of mass 32 kg. its wheels touch the ground at point A and B .1.50m apart and the center of gravity of the bicycle lies vertically above a point half way along the line AB. The Rider sitting on the bicycle is of mass 90 kg and his center of gravity lies vertically above a point on AB=0.50m in front of B find the normal reactions FA and FB which the ground exerts at A and B respectively, when the bicycle is in equilibrium (take g=10cm/s²
To find the normal reactions FA and FB, we need to consider the forces acting on the bicycle in equilibrium.
Let's start by considering only the bicycle itself. The weight of the bicycle can be represented by a downward force acting at its center of gravity (CG). The magnitude of this force is given by:
Fbicycle = mass_bicycle * g
Fbicycle = 32 kg * 10 m/s^2
Fbicycle = 320 N
Since the CG of the bicycle is positioned vertically above a point halfway along AB, we know that the normal reaction at point A is equal to the weight of the bicycle. Thus, FA = Fbicycle = 320 N.
Now, let's consider the rider as well. The weight of the rider can also be represented by a downward force acting at his center of gravity (CG). The magnitude of this force is given by:
Frider = mass_rider * g
Frider = 90 kg * 10 m/s^2
Frider = 900 N
However, the rider's CG is positioned on AB at a point 0.5 m in front of B. Therefore, to find FB, we need to consider the torque caused by the rider's weight about point B. The moment (torque) is given by the equation:
Moment = Force * Distance
Since the bicycle is in equilibrium, we know that the clockwise moment caused by the rider's weight must be balanced by the counter-clockwise moment caused by the bicycle's weight. Therefore, we have:
Fbicycle * (0.5 m) = Frider * (1.5 m - 0.5 m)
320 N * (0.5 m) = 900 N * (1.5 m - 0.5 m)
160 N * 0.5 m = 900 N * 1 m
80 N * m = 900 N * m
Simplifying the equation:
80 N = 900 N
However, this is not a valid equation, which means our assumption must have been incorrect. Therefore, let's assume that the normal reaction at point B, FB, is equal to the weight of both the bicycle and the rider. Thus, FB = Fbicycle + Frider.
FB = 320 N + 900 N
FB = 1220 N
Note that this assumption holds as long as the total weight of the bicycle and rider is not enough to lift the rear wheel off the ground.
Therefore, the normal reactions FA and FB exerted by the ground at points A and B, respectively, when the bicycle is in equilibrium, are:
FA = 320 N
FB = 1220 N
To find the normal reactions FA and FB exerted by the ground at points A and B respectively, we need to consider the equilibrium of forces acting on the bicycle and the rider. We'll consider the forces acting in the vertical direction.
1. First, let's find the total weight of the bicycle and the rider:
Weight of the bicycle = mass of the bicycle × acceleration due to gravity
= 32 kg × 10 m/s² (converting g from cm/s² to m/s²)
= 320 N
Weight of the rider = mass of the rider × acceleration due to gravity
= 90 kg × 10 m/s²
= 900 N
Total weight (bicycle + rider) = 320 N + 900 N
= 1220 N
2. With the given information, we can determine that the center of gravity of the bicycle and the rider lies vertically above a point at a distance of 0.50 m in front of point B. This means that the distance between the center of gravity and point B is 0.50 m.
3. Now, let's consider the equilibrium of forces in the vertical direction.
The forces acting in the vertical direction are:
- The normal reaction FA exerted by the ground at point A
- The normal reaction FB exerted by the ground at point B
- The weight of the bicycle and rider (1220 N)
The sum of the vertical forces should be zero for the bicycle to be in equilibrium.
∑Fy = 0
FA + FB - (weight of bicycle and rider) = 0
Substituting the values:
FA + FB - 1220 N = 0
4. Additionally, we can use the fact that the center of gravity of the bicycle lies vertically above a point halfway along the line AB. The distance between A and B is 1.50 m, and the center of gravity is halfway, so the distance between the center of gravity and A (and B) is 0.75 m.
Using the principle of moments (torques), we can write an equation for the equilibrium of moments around point A:
∑τA = 0
The torques acting are:
- Torque due to FA = FA × distance between A and center of gravity (0.75 m)
- Torque due to FB = FB × distance between B and center of gravity (0.75 m)
- Torque due to the weight of the bicycle and rider = (weight of bicycle and rider) × distance between center of gravity and A (0.50 m)
Setting up the equation:
FA × 0.75 m + FB × 0.75 m - 1220 N × 0.50 m = 0
5. Now we have two equations:
FA + FB - 1220 N = 0 (equation 1)
FA × 0.75 m + FB × 0.75 m - 1220 N × 0.50 m = 0 (equation 2)
We can solve these two equations simultaneously to find the values of FA and FB.
Note: The units used are m for distance and N for force.
To find the normal reactions FA and FB, we need to consider the forces acting on the bicycle and the rider. In this case, there are three forces that contribute to the equilibrium: the weight of the bicycle, the weight of the rider, and the normal reactions at points A and B.
Step 1: Calculate the weight of the bicycle and the rider
The weight of an object is given by the formula: weight = mass × acceleration due to gravity (g).
Weight of the bicycle = mass of the bicycle × g = 32 kg × 10 m/s² = 320 N
Weight of the rider = mass of the rider × g = 90 kg × 10 m/s² = 900 N
Step 2: Calculate the total weight acting on the bicycle and the rider system
The total weight acting on the system is the sum of the weights of the bicycle and the rider.
Total weight = Weight of the bicycle + Weight of the rider
Total weight = 320 N + 900 N = 1220 N
Step 3: Calculate the position of the center of gravity of the bicycle and rider system
The center of gravity is the point at which the weight of an object can be considered to act. Given that the center of gravity of the bicycle lies vertically above a point halfway along the line AB, and the rider's center of gravity lies vertically above a point on AB that is 0.50 m in front of B, we can calculate the distance of the center of gravity from point A.
Distance of the center of gravity from point A (d) = (0.50 m + 1.50 m/2) = 1.25 m
Step 4: Calculate the normal reactions at points A and B
The sum of the vertical forces acting on an object in equilibrium (when not accelerating vertically) is zero. In this case, the sum of the vertical forces at points A and B is equal to the total weight.
Forces at A: FA - Total weight = 0
FA = Total weight = 1220 N
Forces at B: FB - Total weight = 0
FB = Total weight = 1220 N
Therefore, the normal reactions at points A and B are both 1220 N.
In summary,
FA = 1220 N
FB = 1220 N