Two teams of nine members each engage in a tug of war. The first team's members have average masses of 67 kg and exert average forces of 1350 N horizontally. The second team's members have average masses of 72 kg and exert average forces of 1361 N horizontally.
(a) What is the acceleration of the two teams?
(b) What is the tension in the section of rope between the teams?
a) F = ma this is Newton's second law.
For each team you are given F and m
All you have to do is solve for a:
a = F/m That should be easy to do.
b) The tension is easy too, both teams are pulling on the horizontal so it's merely 1361 N - 1350 N = 11 N.
Howdy Jason, this is Michelle! Unfortunately I tried that, as well as just adding the forces together. Neither are the correct answer. Thanks for trying to help tho :).
I think the answer might be 1361 N. The larger force will pull the rope in the direction of that force causing the tension to be 1361 N.
To answer these questions, we'll apply 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 (F = ma).
(a) To find the acceleration of each team, we need to calculate the net force acting on them. In a tug of war, the net force is the difference between the forces exerted by the two teams. Let's calculate it:
Net force on team 1 = Force exerted by team 1 - Force exerted by team 2
= 1350 N - 1361 N
= -11 N
Net force on team 2 = Force exerted by team 2 - Force exerted by team 1
= 1361 N - 1350 N
= 11 N
Since the net force on team 1 and team 2 are equal in magnitude but opposite in direction, the accelerations of both teams will also be equal in magnitude but opposite in direction.
Using Newton's second law, we can find the acceleration:
For team 1:
Force = mass * acceleration
-11 N = 67 kg * acceleration1
acceleration1 = -11 N / 67 kg
acceleration1 ≈ -0.1642 m/s²
For team 2:
Force = mass * acceleration
11 N = 72 kg * acceleration2
acceleration2 = 11 N / 72 kg
acceleration2 ≈ 0.1528 m/s²
Therefore, the acceleration of team 1 is -0.1642 m/s² (directed to the left) and the acceleration of team 2 is approximately 0.1528 m/s² (directed to the right).
(b) To find the tension in the section of rope between the teams, we can consider the rope as a single object. Since the rope is not accelerating in the horizontal direction (there is no net force on the rope), the tension in the rope will be equal to the force exerted by either team.
The tension in the rope is equal to the force exerted by team 1 (1350 N) or team 2 (1361 N). Therefore, the tension in the section of rope between the teams is approximately 1350 N or 1361 N, depending on which team you consider.