a 20 kg traveler's bag is being pulled along the floor with a force of 50 NF. the force is applied on the handle forming an angle of 30 degree with the horizontal. neglecting friction,what is the acceleration og the bag? How much force is exerted by the floor on the bag?
See previous post: Sat, 11-29-14, 11:06 PM.
To find the acceleration of the bag, we will use Newton's second law of motion:
F = m * a
Where:
F = Applied force
m = Mass of the bag
a = Acceleration of the bag
Given:
Mass of the bag (m) = 20 kg
Applied force (F) = 50 N
Since the force is applied at an angle to the horizontal, we need to find the horizontal component of the force (F_horizontal). This can be done using trigonometry:
F_horizontal = F * cos(theta)
Where:
theta = Angle between the applied force and the horizontal axis
F_horizontal = Horizontal component of the applied force
Given:
theta = 30 degrees
Substituting the values, we get:
F_horizontal = 50 N * cos(30 degrees)
F_horizontal = 50 N * 0.866
F_horizontal ≈ 43.3 N
Now, substituting these values into Newton's second law equation:
F_horizontal = m * a
43.3 N = 20 kg * a
Simplifying the equation:
a = 43.3 N / 20 kg
a ≈ 2.165 m/s²
So, the bag has an acceleration of approximately 2.165 m/s².
To find the force exerted by the floor on the bag, we can use Newton's third law of motion, which states that the force exerted by object A on object B is equal in magnitude and opposite in direction to the force exerted by object B on object A.
Therefore, the magnitude of the force exerted by the floor on the bag is also 50 N, but in the opposite direction.
To find the acceleration of the bag, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, we need to resolve the force applied on the handle into its horizontal and vertical components. The horizontal component (F_h) is given by F_h = F * cosθ, and the vertical component (F_v) is given by F_v = F * sinθ, where F is the force applied on the handle and θ is the angle it forms with the horizontal.
Given:
Mass of the bag (m) = 20 kg
Force applied on the handle (F) = 50 N
Angle with the horizontal (θ) = 30 degrees
Calculating the horizontal component of the force:
F_h = F * cosθ
F_h = 50 N * cos(30°)
F_h = 50 N * 0.866 (rounded to three decimal places)
F_h ≈ 43.3 N
Now, since there is no friction involved, the horizontal component of the force will be responsible for the acceleration of the bag. So, we can equate the horizontal component to the product of mass and acceleration:
F_h = m * a
Rearranging the equation to solve for acceleration:
a = F_h / m
a = 43.3 N / 20 kg
a ≈ 2.17 m/s^2 (rounded to two decimal places)
The acceleration of the bag is approximately 2.17 m/s^2.
To find the force exerted by the floor on the bag, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In this case, the force exerted by the floor on the bag (F_floor) will be equal in magnitude but opposite in direction to the horizontal component of the force (F_h).
Thus, the force exerted by the floor on the bag is approximately 43.3 N in the opposite direction.