A firefighter with a weight of 722 N slides down a vertical pole with an acceleration of 3.11 m/s2, directed downward.
(a) What are the magnitude and direction of the vertical force (use up as the positive direction) exerted by the pole on the firefighter?
N
(b) What are the magnitude and direction of the vertical force (use up as the positive direction) exerted by the firefighter on the pole?
N
a) Normally, al objects on earth fall with the same (nearly constant)acceleration (namely 9.81 m/s²).
We see here that the firefighter only slides down the pole with an acceleration of 3.11 m/s². This means, that in order to slow down the firefighter, there must be an upward acceleration of 6.70 m/s² which results from the force of the pole acting on the firefighter.
Now, according to Newtons first law F = m.a (Force equals mass times acceleration)
So, since we have our upward acceleration and since we have the mass of the firefighter (m=722N/(9.81 m/s²) = 73.6 kg), we can calculate the force of the pole acting on the firefighter as follows:
F= m.a = 73,6 kg . 6.70 m/s² = 493.12 N
This force acts in the upward direction.
To solve this problem, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
(a) First, we need to calculate the net force acting on the firefighter. The net force is given by the product of the mass (converted from weight) and the acceleration:
Net force = mass x acceleration
To convert the given weight (722 N) to mass, we divide it by the acceleration due to gravity, which is approximately 9.8 m/s^2:
Mass = Weight / Acceleration due to gravity
Mass = 722 N / 9.8 m/s^2
Substituting the values, we have:
Mass = 73.78 kg
Therefore, the net force acting on the firefighter is:
Net force = 73.78 kg x 3.11 m/s^2
(b) The vertical force exerted by the firefighter on the pole is equal in magnitude and opposite in direction to the force exerted by the pole on the firefighter. Therefore, it is also equal to the net force calculated in part (a):
Magnitude of the vertical force exerted by the firefighter on the pole = Magnitude of the net force = 73.78 kg x 3.11 m/s^2
To determine the direction of the force, we need to consider the sign of the net force. Since the firefighter is sliding downward, the net force is downward, and the vertical force exerted by the firefighter on the pole is also downward.
Therefore:
(a) The magnitude of the vertical force exerted by the pole on the firefighter is equal to the net force, which is 73.78 kg x 3.11 m/s^2.
(b) The magnitude of the vertical force exerted by the firefighter on the pole is also equal to the net force, which is 73.78 kg x 3.11 m/s^2. The direction of this force is downward.
To find the magnitude and direction of the vertical forces exerted by the pole on the firefighter and the firefighter on the pole, we can use Newton's second law of motion, which states that the sum of all forces acting on an object is equal to the product of its mass and acceleration.
Given:
Weight of the firefighter (force due to gravity) = 722 N
Acceleration = 3.11 m/s^2 (directed downward)
(a) Magnitude and direction of the vertical force exerted by the pole on the firefighter:
The vertical force exerted by the pole on the firefighter can be calculated by subtracting the force due to gravity from the net force acting on the firefighter.
Net force = mass x acceleration
Since the mass is not given, we can use the equation: Weight = mass x gravitational acceleration
722 N = mass x 9.8 m/s^2 (gravitational acceleration)
mass = 722 N / 9.8 m/s^2 = 73.88 kg
Net force = mass x acceleration
= 73.88 kg x 3.11 m/s^2
= 229.37 N (upward)
Therefore, the magnitude of the vertical force exerted by the pole on the firefighter is 229.37 N, directed upward.
(b) Magnitude and direction of the vertical force exerted by the firefighter on the pole:
According to Newton's third law of motion, the force exerted by the firefighter on the pole is equal in magnitude but opposite in direction to the force exerted by the pole on the firefighter.
Hence, the magnitude of the vertical force exerted by the firefighter on the pole is also 229.37 N, directed downward.