A construction worker hoists himself up a building with the apparatus shown in the figure. A rope is attached to a chair and passes through a massless pulley that can turn without friction. The worker pulls on the free end to lift himself. Suppose that the combined mass of the chair and the man is 100 kg. With what force must the man pull down on the free end of the rope so that his upward acceleration is 0.940 m/s2?
net force=mass*a
force on left-forceonright=mass*a
force on left -100*9.8=100*.940
To find the force with which the man must pull down on the free end of the rope, we can use Newton's second law of motion.
1. Identify the relevant forces:
- The gravitational force acting on the man and the chair (weight)
- The tension force in the rope
2. Determine the forces:
- The gravitational force on the man and the chair is given by:
F_gravity = mass * gravity
F_gravity = 100 kg * 9.8 m/s^2
F_gravity = 980 N (upward)
- The tension force in the rope pulls the man and the chair upward. Let's denote it as T.
3. Apply Newton's second law of motion:
The net force acting on the system (man and chair) is given by:
F_net = m * a
Since the system is moving upward, the net force is the difference between the tension force and the gravitational force:
F_net = T - F_gravity
Rearranging the equation, we get:
T = F_gravity + F_net
T = F_gravity + (m * a)
4. Substitute the given values:
- F_gravity = 980 N (upward)
- m = 100 kg
- a = 0.940 m/s^2
T = 980 N + (100 kg * 0.940 m/s^2)
5. Calculate the tension force:
T = 980 N + 94 N
T = 1074 N
Therefore, the man must pull down on the free end of the rope with a force of 1074 N in order to achieve an upward acceleration of 0.940 m/s^2.