When a man in a cage whose weight is 1000N pulls a rope that is attached on a cage around a fixed pulley,the force he exerts on the floor of the cage is 450N ,if the cage weigh 250N ,find the acceleration
I don't get the picture...a man in a cage pulls a rope attached to the cage..
Picture it like this:the rope is attached on top of the cage then from there it passes around a fixed pulley so the man in the cage is pulling the rope down from the pulley.PLEASE HELP ME Bobpursley
The weight of cage + man is 1250 N and their combined mass is
Mtotal = Wtotal/g = 127.6 kg.
The man and cage accelerate at the same rate, a. Let F be the rope tension force. Consider all forces on the man, including weight and the upward floor force. The man's mass is 102 kg
F - Wman + 450 = F - 550 = 102*a
Now consider all exernal forces on man and cage together:
F -1250 = 127.6 a
Combining the two results in
700 = -25.6 a
a = -0.0366 m/s^2
I think u have to also take
T-f-Wchair=McA
U did a mistake
Wtotal/g=127.55
To find the acceleration, 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:
F_net = m * a
In this case, the net force is the force exerted on the floor of the cage, which is 450N. The mass of the object is the total weight of the man and the cage, which is (1000N + 250N) / 9.8 (using the conversion factor of 9.8 N/kg for weight to mass). Let's calculate:
F_net = m * a
450N = [(1000N + 250N) / 9.8] * a
Now, let's solve for acceleration:
450N = (1250N / 9.8) * a
450N * 9.8 = 1250N * a
4410N = 1250N * a
To solve for acceleration, divide both sides of the equation by 1250N:
a = 4410N / 1250N
Simplifying:
a = 3.528 m/s^2
Therefore, the acceleration of the cage is approximately 3.528 m/s^2.