Two people pull as hard as they can on ropes attached to a 200 kg boat. If they pull in the same direction, the boat has an acceleration of 1.44 m/s2 to the right. If they pull in opposite directions, the boat has an acceleration of 0.518 m/s2 to the left. What is the force exerted by each person on the boat? (Disregard any other forces on the boat.)
Smaller N
Larger N
n+N=200*1.44
N-n=200(.518)
add the equations
2N=200(1.44 +.518)
solve for N
N=100(1.96)=196N
n=200*1.44-195=288-196=92 N
To find the force exerted by each person on the boat, we can 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.
When the two people are pulling in the same direction, the net force on the boat is the sum of the forces exerted by each person. Let's call the force exerted by the smaller person F1 and the force exerted by the larger person F2.
Using Newton's second law, we can write the equation:
F1 + F2 = m × a
where m is the mass of the boat and a is the acceleration.
Given that the mass of the boat (m) is 200 kg and the acceleration (a) is 1.44 m/s², we can substitute these values into the equation:
F1 + F2 = 200 kg × 1.44 m/s²
F1 + F2 = 288 N
When the two people are pulling in opposite directions, the net force on the boat is the difference between the forces exerted by each person. Let's call the force exerted by the smaller person F3, which is in the opposite direction to the motion, and the force exerted by the larger person F4.
Using Newton's second law again, we can write the equation:
F4 - F3 = m × a
Using the same mass (m) and a new acceleration value, which is opposite in direction (-0.518 m/s²), we can substitute these values into the equation:
F4 - F3 = 200 kg × (-0.518 m/s²)
F4 - F3 = -103.6 N
Now we have a system of two equations:
F1 + F2 = 288 N
F4 - F3 = -103.6 N
To solve this system, we'll use the method of substitution. Rearrange one equation to express a variable in terms of the other and substitute it into the second equation.
From the first equation, we can express F1 as:
F1 = 288 N - F2
Now substitute this expression for F1 into the second equation:
F4 - F3 = -103.6 N
F4 - (288 N - F2) = -103.6 N
F4 + F2 - 288 N = -103.6 N
Rearrange the equation:
F4 + F2 = 184.4 N
Now we have two equations:
F1 + F2 = 288 N
F4 + F2 = 184.4 N
To solve this system, we can add the two equations:
(F1 + F2) + (F4 + F2) = 288 N + 184.4 N
F1 + 2F2 + F4 = 472.4 N
Now we can substitute the value of F4 + F2 from the previous step:
F1 + 2F2 + 184.4 N = 472.4 N
Rearrange the equation:
F1 + 2F2 = 472.4 N - 184.4 N
F1 + 2F2 = 288 N
Now we have a simplified equation:
F1 + 2F2 = 288 N
Substitute the value of F1 from the first equation:
(288 N - F2) + 2F2 = 288 N
Simplify the equation:
288 N - F2 + 2F2 = 288 N
F2 = 0 N
Now substitute the value of F2 back into the first equation:
F1 + 0 N = 288 N
F1 = 288 N
So, the force exerted by the smaller person (F1) is 288 N, and the force exerted by the larger person (F2) is also 288 N.
To solve this problem, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
Let's say the force exerted by the smaller person is F1 (in Newtons) and the force exerted by the larger person is F2 (in Newtons).
When they pull in the same direction, the net force acting on the boat is the sum of the forces exerted by each person:
F_net = F1 + F2
According to Newton's second law, F_net = m*a, where m is the mass of the boat and a is its acceleration.
So, in this case, we have F1 + F2 = m*a, which can be written as F1 + F2 = 200 kg * 1.44 m/s^2.
Similarly, when they pull in opposite directions, the net force is the difference between the forces exerted by each person:
F_net = F1 - F2
According to Newton's second law, F_net = m*a, where m is the mass of the boat and a is its acceleration.
So, in this case, we have F1 - F2 = 200 kg * (-0.518 m/s^2).
Now we have a system of two equations with two unknowns (F1 and F2). We can solve this system of equations to find the values of F1 and F2.
Using the equations F1 + F2 = 200 kg * 1.44 m/s^2 and F1 - F2 = 200 kg * (-0.518 m/s^2), we can add the two equations together:
(F1 + F2) + (F1 - F2) = 200 kg * 1.44 m/s^2 + 200 kg * (-0.518 m/s^2)
This simplifies to:
2F1 = 200 kg * (1.44 m/s^2 - 0.518 m/s^2)
Simplifying further:
2F1 = 200 kg * 0.922 m/s^2
F1 = (200 kg * 0.922 m/s^2) / 2
F1 = 92.2 N
Now that we have the value of F1, we can substitute it back into either of the original equations to solve for F2.
Using the equation F1 + F2 = 200 kg * 1.44 m/s^2:
92.2 N + F2 = 200 kg * 1.44 m/s^2
F2 = 200 kg * 1.44 m/s^2 - 92.2 N
F2 = 288 N - 92.2 N
F2 = 195.8 N
So, the force exerted by the smaller person is 92.2 N and the force exerted by the larger person is 195.8 N.