Solving Two Problems with Pulleys and Cords: Tensions and Distances

[GingerBread27]

In Figure 5-50, three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 22.0 kg, mB = 40.0 kg, mC = 18.0 kg.

Fig. 5-50
http://www.webassign.net/hrw/W0084-N.jpg

(a) When the assembly is released from rest, what is the tension in the cord that connects boxes B and C?
N
(b) How far does box A move in the first 0.250 s (assuming it does not reach the pulley)?
m

and the other is

Figure 5-55 shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 7.00 kg, mB = 8.00 kg, mC = 10.5 kg. When the blocks are released, what is the tension in the cord at the right?
N

http://www.webassign.net/hrw/W0080-N.jpg

 

[Pyrrhus]

What have you tried? What are the forces acting on each of the blocks on the x-axis and y-axis?

 

[M98Ranger]

Pulleys and cords are a great tool for solving mechanical problems, as they allow us to easily manipulate forces and distances. In the first scenario, we have three ballot boxes connected by cords, with one pulley having negligible friction and mass. We are given the masses of the boxes and asked to find the tension in the cord connecting boxes B and C, as well as the distance box A will move in the first 0.250 seconds.

To solve for the tension in the cord, we can use the equation T1 = T2, where T1 is the tension in the cord connecting boxes B and C and T2 is the tension in the cord connecting box A to the pulley. We know that the weight of the boxes is equal to the tension in the cords, so we can set up the following equation:

mBg = mCg + T1

Substituting in the given values, we get:

(40.0 kg)(9.8 m/s^2) = (18.0 kg)(9.8 m/s^2) + T1

Solving for T1, we get a tension of 196 N in the cord connecting boxes B and C.

To find the distance box A will move in the first 0.250 seconds, we can use the equation d = 1/2at^2, where d is the distance, a is the acceleration, and t is the time. We know that the acceleration is equal to the tension in the cord divided by the mass of box A, so we can set up the following equation:

d = 1/2(T2/mA)t^2

Substituting in the given values, we get:

d = 1/2[(22.0 kg)(9.8 m/s^2)/22.0 kg](0.250 s)^2

Solving for d, we get a distance of 0.306 m for box A in the first 0.250 seconds.

In the second scenario, we have three blocks attached by cords that loop over frictionless pulleys. We are given the masses of the blocks and asked to find the tension in the cord at the right when the blocks are released.

To solve for the tension in the cord, we can again use the equation T1 = T2, where T1 is the tension in the cord connecting blocks B and C and T2 is the tension in the cord connecting