Two Force Acting Simultaneously

  • #1
BurpHa
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Homework Statement
A 10.0 kg object is launched vertically into the air with an initial velocity of 50.0 m/s. In addition to the force of gravity there is a frictional force which is proportional to velocity according to ##f_y=-bv_y##; note that this frictional force is negative (down) when the object is moving up, but positive (up) when the object is moving down.

(a) Numerically generate distance-time graphs for the object,
using ##b=0## but use several different step sizes for ##\Delta t##, such as 1.0 s, 0.1 s, 0.01 s, and 0.001 s. Show the results on a single graph. How does the highest point vary with the step size?

(b) Numerically generate distance-time graphs for the object,
using a step size of ##\Delta t=0.01 s##. Now, however, try non-zero
values for ##b##, such as 0.1 ##N* s/m## 0.5 ##N* s/m## 1.0 ##N* s/m##
5.0 ##N* s/m## and 10.0 ##N* s/m## How does the highest point vary
with ##b##? What do you notice about the shape of the graphs as ##b##
increases?
Relevant Equations
##V_f=V_i-at##
I could understand the problem perfectly; however, I do not know how to construct the problem. The problem states that two forces are acting simultaneously on the object, but how could I represent that fact mathematically?

I really want to solve it, but I am facing this roadblock, so please only give me hints.

Thank you.
 
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  • #2
BurpHa said:
Homework Statement: A 10.0 kg object is launched vertically into the air with an initial velocity of 50.0 m/s. In addition to the force of gravity there is a frictional force which is proportional to velocity according to ##f_y=-bv_y##; note that this frictional force is negative (down) when the object is moving up, but positive (up) when the object is moving down.

(a) Numerically generate distance-time graphs for the object,
using ##b=0## but use several different step sizes for ##\Delta t##, such as 1.0 s, 0.1 s, 0.01 s, and 0.001 s. Show the results on a single graph. How does the highest point vary with the step size?

(b) Numerically generate distance-time graphs for the object,
using a step size of ##\Delta t=0.01 s##. Now, however, try non-zero
values for ##b##, such as 0.1 ##N* s/m## 0.5 ##N* s/m## 1.0 ##N* s/m##
5.0 ##N* s/m## and 10.0 ##N* s/m## How does the highest point vary
with ##b##? What do you notice about the shape of the graphs as ##b##
increases?
Relevant Equations: ##V_f=V_i-at##

I could understand the problem perfectly; however, I do not know how to construct the problem. The problem states that two forces are acting simultaneously on the object, but how could I represent that fact mathematically?

I really want to solve it, but I am facing this roadblock, so please only give me hints.

Thank you.
Forces add like vectors - although in this case, we have motion in only one dimension. If there are two forces acting on an object then it's the resultant force (sum of all forces) that applies in Newton's second law:
$$\vec F = m\vec a, \ \text{where} \ \vec F = \vec F_1 + \vec F_2$$
 
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  • #3
How "numerical" does your teacher want your solution to be?

Would solving the differential equation analytically and then plotting the solutions be acceptable?

https://tutorial.math.lamar.edu/classes/de/modeling.aspx If so check out example 4 which does exactly what was mentioned in post #2.
 
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  • #4
PeroK said:
Forces add like vectors - although in this case, we have motion in only one dimension. If there are two forces acting on an object then it's the resultant force (sum of all forces) that applies in Newton's second law:
$$\vec F = m\vec a, \ \text{where} \ \vec F = \vec F_1 + \vec F_2$$
But the friction force depends on the velocity, which I do not know how to formulate.
 
  • #5
PhDeezNutz said:
How "numerical" does your teacher want your solution to be?

Would solving the differential equation analytically and then plotting the solutions be acceptable?

https://tutorial.math.lamar.edu/classes/de/modeling.aspx If so check out example 4 which does exactly what was mentioned in post #2.
Thank you! I think this is what I need.
 
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  • #6
BurpHa said:
But the friction force depends on the velocity, which I do not know how to formulate.
Have you heard of differential equations?
 
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  • #7
BurpHa said:
But the friction force depends on the velocity, which I do not know how to formulate.
So, for example, with one-dimensional motion in the y-direction, Newton's second law becomes:
$$F_y = ma_y = m\frac{d^2 y}{dt^2}$$
 
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