As shown in the figure below, a stick of length
L = 0.330 m
and mass
m = 0.205 kg
is in contact with a rough floor at one end and a frictionless bowling ball (diameter
d = 21.50 cm)
at some other point such that the angle between the stick and the floor is
θ = 30°.
Determine the following.
(a) magnitude of the force exerted on the stick by the bowling ball
N
(b) horizontal component of the force exerted on the stick by the floor
N
(c) vertical component of the force exerted on the stick by the floor
N
To determine the magnitude of the force exerted on the stick by the bowling ball (a), the horizontal component of the force exerted on the stick by the floor (b), and the vertical component of the force exerted on the stick by the floor (c), we can use the principles of equilibrium and solve for the unknown forces.
First, let's analyze the forces acting on the stick. We have the gravitational force acting downward at the center of mass of the stick, which can be represented by the weight W = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
(a) To determine the magnitude of the force exerted on the stick by the bowling ball, we can consider the torque equation. By summing up the torques about the contact point between the stick and the floor, where the normal force from the floor acts, we can find the equilibrium condition.
The torque equation can be written as:
τ = F * r * sin(θ)
Where τ is the torque, F is the force, r is the perpendicular distance from the point of application of the force to the pivot, and θ is the angle between the force vector and the line drawn from the pivot to the point of application of the force.
In this case, the torque caused by the weight of the stick can be disregarded since it is acting along the line of action of the force from the floor. The only other force that can generate torque is the force exerted by the bowling ball.
The torque exerted by the bowling ball is τ = F_b * L * sin(θ) since the perpendicular distance from the pivot to the point of contact with the bowling ball is L.
The equilibrium condition states that the sum of the torques is zero:
0 = F_b * L * sin(θ)
Rearranging the equation, we can solve for F_b:
F_b = 0 / (L * sin(θ))
Since sin(θ) is zero, this means that the magnitude of the force exerted by the bowling ball is zero. Therefore, the magnitude of the force exerted on the stick by the bowling ball is 0 N.
(b) To determine the horizontal component of the force exerted on the stick by the floor, we need to consider the horizontal equilibrium of forces.
Since the bowling ball does not exert any horizontal force on the stick, the horizontal component of the force exerted on the stick by the floor must counterbalance the horizontal component of the gravitational force. This can be written as:
F_floor_horizontal = W * sin(θ)
F_floor_horizontal = m * g * sin(θ)
Substituting the known values, we can calculate the horizontal component of the force exerted on the stick by the floor (b).
(c) To determine the vertical component of the force exerted on the stick by the floor, we need to consider the vertical equilibrium of forces.
The vertical component of the force exerted on the stick by the floor must counterbalance the vertical component of the gravitational force plus the vertical component of the force exerted by the bowling ball. This can be written as:
F_floor_vertical = W * cos(θ) + F_b
F_floor_vertical = m * g * cos(θ) + 0
Substituting the known values, we can calculate the vertical component of the force exerted on the stick by the floor (c).
To solve this problem, we can break down the forces acting on the stick and solve for each component separately.
Step 1: Draw a free body diagram and label the forces acting on the stick.
Let's break down the forces acting on the stick as follows:
- The weight of the stick (mg) acting vertically downwards.
- Normal force from the floor (N_f) acting upwards and perpendicular to the floor.
- Force of friction (F_f) acting horizontally in the opposite direction of motion.
Step 2: Calculate the weight of the stick.
The weight of the stick can be calculated using the formula:
Weight (W) = mass (m) * acceleration due to gravity (g)
Given:
Mass (m) = 0.205 kg
Acceleration due to gravity (g) = 9.8 m/s^2
W = 0.205 kg * 9.8 m/s^2
W = 2.009 N
Step 3: Calculate the normal force from the floor.
Since the stick is at an angle to the floor, we need to resolve the weight vector into vertical and horizontal components. The vertical component of the weight will be balanced by the normal force from the floor.
Vertical Component of Weight (W_v) = W * sin(θ)
W_v = 2.009 N * sin(30°)
W_v = 1.0045 N
Therefore, the normal force from the floor is 1.0045 N.
Step 4: Calculate the horizontal component of the weight.
The horizontal component of the weight will be balanced by the force of friction from the floor.
Horizontal Component of Weight (W_h) = W * cos(θ)
W_h = 2.009 N * cos(30°)
W_h = 1.7417 N
Therefore, the horizontal component of the weight is 1.7417 N.
Step 5: Calculate the force exerted on the stick by the bowling ball.
Since the stick is in contact with a frictionless bowling ball, there is no horizontal force exerted by the ball. The only vertical force exerted by the ball is its weight, which is balanced by the normal force from the floor.
Therefore, the magnitude of the force exerted on the stick by the bowling ball is equal to the vertical component of the weight of the stick, which is 1.0045 N.
Step 6: Calculate the vertical component of the force exerted on the stick by the floor.
Since the stick is at an angle to the floor, the vertical component of the force exerted by the floor is equal to the vertical component of the weight of the stick, which is 1.0045 N.
Therefore, the vertical component of the force exerted on the stick by the floor is 1.0045 N.
Step 7: Calculate the horizontal component of the force exerted on the stick by the floor.
Since the stick is at an angle to the floor, the horizontal component of the force exerted by the floor is equal to the horizontal component of the weight of the stick, which is 1.7417 N.
Therefore, the horizontal component of the force exerted on the stick by the floor is 1.7417 N.
To summarise:
(a) The magnitude of the force exerted on the stick by the bowling ball is 1.0045 N.
(b) The horizontal component of the force exerted on the stick by the floor is 1.7417 N.
(c) The vertical component of the force exerted on the stick by the floor is 1.0045 N.