Dynamics Prob force is a function of time

Click For Summary
SUMMARY

The discussion focuses on calculating the time required for a block, initially falling at a velocity \( v_0 \), to come to a stop when a motor applies a force \( F(t) \). The net force acting on the block is defined as \( F_{\textrm{net}} = mg - F(t) \), leading to the acceleration \( a(t) = \frac{F_{\textrm{net}}}{m} \). The integral equation \( v_0 + gt_s = \frac{1}{m}\int_0^{t_s} F(t)\, dt \) is derived to express the relationship between the initial velocity, gravitational acceleration, and the applied force over time. The solution requires knowledge of the functional form of \( F(t) \) to proceed further.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Basic calculus, specifically integration techniques
  • Familiarity with kinematic equations
  • Knowledge of force dynamics in mechanical systems
NEXT STEPS
  • Research the functional forms of force functions \( F(t) \) in mechanical systems
  • Study the application of integrals in physics, particularly in motion equations
  • Explore examples of deceleration scenarios in physics problems
  • Learn about the relationship between mass, force, and acceleration in dynamic systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the dynamics of forces acting on moving objects.

fKnight
Messages
2
Reaction score
0

Homework Statement


A block is attached to a rope through a single pulley connected to a motor pulling with a force F(t), t in seconds, when the block is falling at a specific velocity > 0 the motors is applied to stop, find the time it takes for v=0.

neglect mass of pulley & rope.

Homework Equations


The Attempt at a Solution


Is this as simple as setting F(t)=m*g? or would I need to set the forces in the y direction = to mass * accel(in y) then take the integral of dv from v to 0 = the integral of a dt from 0 to t. then solve for t?
 
Physics news on Phys.org
The motor isn't turned on until the block is falling at some rate, which we'll call the initial velocity v0. So, given that the block is initally falling at that speed, how long does it take the motor to decelerate it? Let the time the motor is first turned on be t = 0. Let the stopping time, which we are trying to solve for, be ts. Well, the change in velocity in that time interval is given by the integral of the acceleration:

v(t_s) = v_0 + \int_0^{t_s} a(t)\, dt​

Taking downward to be the positive direction, we know that a(t) = Fnet/m, where m is the mass of the block, and Fnet is the net force on it:

a(t) = \frac{F_{\textrm{net}}}{m} = \frac{mg - F(t)}{m}​

Since the speed at the stopping time is zero, the integral equation becomes:

\Delta v = -v_0 = \int_0^{t_s} g - \frac{F(t)}{m} \, dt​

Eventually you get:

v_0 + gt_s = \frac{1}{m}\int_0^{t_s} F(t)\, dt​

To be honest, I have no idea what to do with that without knowing the functional form of F(t)
 
Nevermind, I think I figured it out.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
8K
Replies
7
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K