- #1

- 60

- 0

## Homework Statement

a 5 kg block with a 10cm length sits at the top of a frictionless, 6.0m 12kg uniform ramp. This ramp is held up by a rope of tension T at a 35* angle to the ground. The block is released and allowed to accelerate. What is the rate at which the tension in the supporting rope is decreasing when the block has traveled for 0.5s

Crappy Sketch of the senario

## Homework Equations

Related Rates (I believe)

Fparallel = FgSinθ

Fperpendicular = FgCosθ

Torque = FdSinθ

Kinematics formulas

## The Attempt at a Solution

To find the rate at which tension is decreasing, youd need to find tension at both points. To find tension at point time= 0.5s, you have to calculate how far along the block has moved.

Point 1) Top of the rope (time=0)

Illustration for the top of the ramp

Since the 5kg mass is 10cm long, the center of mass is in the middle (at 5cm)

The toques clockwise and counterclockwise (FdSinθ) are:

Tcw = Tccw

T * 6 * Sin 90 = (49 * (6-0.05) * Sin 35) + (117.6 * 3 *Sin35)

T = 61.6 N

Point 2) Down the rope (time = 0.5)

Fnet = Fparallel (as there is no friction force)

= MgSinθ

a = (9.8)Sin35

a = 5.621 m/s^2

Vo = 0

Vf = X

t = 0.5

d = ?

a = 5.621

d=VoT+1/2at^2

d= (0) + 1/2(5.621)(.5)^2

d = 0.703 m

Illustration for the Torques

Since the center of mass is in the center of the 5kg block, the point that will rotate is (0.10/2) + (0.703), correct?

Following that logic, the Torques Clockwise and CounterClockwise (FdSinθ) are:

Tcw Tccw

T*6*Sin90 = (49*(6-.753)*Sin35) + (117.6 * 3 * Sin35)

T = 58.3 N

Now heres where Im stuck. I could calculate the Tension when the block is at the top, and use the formulate rate = ΔTorque/ΔTime.

Rate = (58.3 - 61.6) / 0.5s

rate = -6.6 N / s

This would give me a logical answer and I think it would work fine. However, I have a feeling i'm supposed to use related rates to solve this. Any help would be amazing.