How to Analyse Force as a Function of Time

In the previous post ( Dynamic Effects of Linear Motion) we considered the motion of a body (such as a vehicle or component in a machine) moving in a straight line when the acceleration was changing during the motion. 

We shall now consider the case where the force is changing with time.

In engineering, understanding how forces act on objects is fundamental to designing safe and efficient systems. While many problems involve constant forces, real-world applications often require analysing situations where force varies with time. This is known as force as a function of time, or F(t).

Time-dependent forces appear in many engineering systems. For example, during a car crash, the impact force rapidly increases and then decreases over a short time. In machinery, motors and actuators apply forces that ramp up or fluctuate based on programmed input or load conditions. In structural engineering, dynamic loads like wind, waves, or earthquakes also produce forces that change with time.

When force is a function of time, engineers must use tools from calculus and differential equations to predict the resulting motion of objects. Newton’s Second Law, F(t) = m⋅a(t) connects force to acceleration, and by integrating over time, engineers can determine how velocity and position change under varying forces.

Consider a body of mass, m, moving under the influence of a force, F, which is a function of time, t.

F = F(t)

Using Newton’s second law, F = ma, gives:

ma = F(t)

Where

a = ^dv⁄_dt

Therefore

m ^dv⁄_dt = F(t)

Let us know integrate with respect to time:

m ∫ ^dv⁄_dt dt = ∫ F(t) dt

∫ ^dv⁄_dt dt = v

mv = ∫ F(t) dt + c

where c is an arbitrary constant of integration.

Example 1

An engineering component, P, of mass 2kg is moving in a straight line within the machine it is part of. P is initially at rest at a point O on the line of motion. The force acting on P has magnitude (3t² + e^t) N at time t seconds and acts in the direction OP. Calculate:

The speed of P at time t = 2 seconds

The distance travelled at time t = 4 seconds

Using Newton’s second law, F = ma, with F = (3t² + e^t) N and m = 2 kg gives:

F = ma

3t² + e^t = 2a

Using

a = ^dv⁄_dt

Gives

2 ^dv⁄_dt = 3t² + e^t

If we now integrate this with respect to t:

2 ∫ ^dv⁄_dt = ∫ (3t²+e^t) dt

2v = ³⁄₃ (t³) + e^t +c  

2v = t³ + e^t + c

Now, we know from the information given in the question that v = 0 when t = 0 (initially at rest) and so we can substitute these values into this equation to find the value of c:

2(0) = (0)³+ e⁰ + c

0 = 0 +1 + c

c = -1

Hence:

2v = t³ + e^t -1

Finally, if we substitute the value for t = 2:

2v = 2³ + e² -1

v = ^14.4⁄₂ = 7.2 m/s

Now, we turn attention to the distance, x. From above, we know that:

2v = t³ + e^t -1

And we also know that

v = ^dx⁄_dt

Therefore:

2 ^dx⁄_dt = t³ + et -1

As before, let us integrate this with respect to time, t:

2 ∫ ^dx⁄_dt dt = t³ + e^t -1

2x = ¹⁄₄ (t⁴)+ e^t – t + k

where k is some arbitrary constant of integration.

As x = 0 when t = 0, we can substitute this into the above equation to give:

0 = 0 + e⁰ + k

Where

e⁰ = 1

k = -1

So the equation for displacement, x, becomes:

2x = ¹⁄₄ t⁴ + e^t – t -1

We are interested in the value of displacement, x, when t = 4:

2x = ¹⁄₄ (4)⁴ + e⁴ – 4 -1

2x = 64 + e⁴ – 5 = 113.6

x = 56.8 m

Example 2

This example explores the motion of an engineering component subject to a time-dependent force. The component, denoted P, has a mass of 3 kg and moves in a straight line within the machine it is part of. At time t = 0, the component is initially at rest at point O, which we take as the origin of its motion.

The force acting on the component varies with time and is given by:

F(t) = 2t² + e^t (in newtons)

Since this force acts along the direction of motion, it causes the component to accelerate. We will apply Newton’s Second Law to determine how this force affects the component’s velocity and displacement over time.

Mass m = 3 kg

Initial condition: at rest at point O, so v(0) = 0

Force: F(t) = 2t² + e^t (Newtons)

We are to find:

Velocity as a function of time v(t)

Velocity at a specific time, e.g. t = 2

Displacement after a given time, e.g. t = 2

F(t) = m ^dv⁄_dt

Substitute in the known values:

2t² + e^t = 3 ^dv⁄_dt

^dv⁄_dt = ^{(2t2 + et ) ⁄₃}

Now integrate both sides:

∫dv = ∫(^2t2⁄₃ +^et⁄₃)dt

∫ dv = ²⁄₃ ∫ (t²dt + ¹⁄₃ ∫ e^t )dt

v(t) =(²⁄₃).^t3 ⁄₃ + ¹⁄₃e^t + c

At t = 0, v(0) = 0

0 = (²⁄₃).⁰³ ⁄₃ + ¹⁄₃.(e⁰) + c

0 = 0 + ¹⁄₃ + c

c = ^-1⁄₃

So the expression becomes

v(t) = (²⁄₉) t³ + ¹⁄₃ e^{t -1}⁄₃

When t = 2

v(2) = (²⁄₉ )2³ + ¹⁄₃e^{2 -1}⁄₃

v(2) = 3.907 m/s

To find the displacement , we need to integrate v(t)

x(t) = ∫ v(t) dt = ∫ (²⁄₉) t³ + ¹⁄₃ e^{t -1}⁄₃ )dt

x(t) = ²⁄₉ ∫ t³.dt +¹⁄₃ ∫ e^t.dt – ¹⁄₃ ∫ 1.dt

x(t) = ¹⁄₁₈t⁴ + ¹⁄₃e^t –¹⁄₃t + c

Since x(0) = 0 when t = 0

0 = 0 + ¹⁄₃ – 0 +c

c = ^-1⁄₃

x(t) = ¹⁄₁₈t⁴ + ¹⁄₃e^t –¹⁄₃t – ¹⁄₃

Now substitute t = 2

x(2) = ¹⁶⁄₁₈ + ¹⁄₃ (3.907) – ²⁄₃ – ¹⁄₃ = 1.1912m

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