Force–time graphs Knowledge-Based Study

Study notes

1. Force and Momentum Over Time

   In classical mechanics, Newton's second law defines net force as the rate of change of linear momentum:

   $$F_{\text{net}} = \frac{\Delta p}{\Delta t}$$

   When this equation is rearranged, we define the quantity known as impulse ($J$ or $I$):

   $$I = F_{\text{net}} \Delta t = \Delta p = m v_f - m v_i$$

   This relationship is called the impulse-momentum theorem. When a constant force acts on an object of mass $m$ over an elapsed time $\Delta t$, calculating the impulse is straightforward multiplication. However, in most real-world interactions—such as a collision, an athletic strike, or rocket propulsion—the applied force varies continuously with time. To analyze such scenarios, we rely on the force–time graph.

   Geometric Meaning of Area

   When force is plotted on the vertical axis ($y$-axis) against time on the horizontal axis ($x$-axis), the geometric area bounded by the curve and the time axis represents the total impulse delivered to the object.

   • Constant Force: The graph forms a rectangle of height $F$ and width $\Delta t$. The area is $\text{Area} = F \Delta t = I$.
   • Linearly Varying Force: Collisions often feature a force that rises linearly to a peak value $F_{\text{max}}$ and falls linearly back to zero. The area forms a triangle: $\text{Area} = \frac{1}{2} F_{\text{max}} \Delta t = I$.
   • Arbitrary Complex Shapes: For piecewise linear graphs (trapezoids, combinations of triangles and rectangles), the total impulse is the algebraic sum of the individual geometric areas. In calculus terms, $I = \int_{t_1}^{t_2} F(t) \, dt$.

   Because impulse equals the change in linear momentum, determining the area directly yields $\Delta p$:

   $$\text{Area under } F\text{--}t \text{ curve} = I = \Delta p = m(v_f - v_i)$$

   Sign Conventions and Net Impulse

   Force and momentum are both vector quantities. Consequently, the orientation of the force on a force–time graph conveys crucial directional information:

   • Positive Area (Above the Time Axis): Indicates a force acting in the designated positive coordinate direction. This imparts a positive impulse, accelerating an object forward or decelerating an object moving in the negative direction.
   • Negative Area (Below the Time Axis): Indicates a force acting in the negative coordinate direction. This imparts a negative impulse, decreasing positive momentum or increasing negative momentum.

   To determine the net change in momentum over a time interval spanning both positive and negative regions, calculate the algebraic sum of the areas:

   $$I_{\text{net}} = \text{Area}_{\text{above}} - |\text{Area}_{\text{below}}| = \Delta p_{\text{net}}$$

   If the positive and negative areas are equal over a certain duration, the net impulse is zero, meaning the object's final momentum equals its initial momentum.

   Step-by-Step Problem Solving

   Consider a cart of mass $m = 2\text{ kg}$ initially traveling along a straight track with an initial velocity $v_i = +3\text{ m/s}$. A variable horizontal force acts on the cart according to the following profile:

   • From $t = 0\text{ s}$ to $t = 4\text{ s}$, the force increases linearly from $0\text{ N}$ to $+10\text{ N}$.
   • From $t = 4\text{ s}$ to $t = 6\text{ s}$, the force remains constant at $+10\text{ N}$.
   • From $t = 6\text{ s}$ to $t = 8\text{ s}$, the force switches direction and remains constant at $-5\text{ N}$.

   1. Calculate the Impulse for Each Segment

   Segment 1 (Triangle, 0 to 4 s):
   $I_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4\text{ s} \times 10\text{ N} = +20\text{ N}\cdot\text{s}$

   Segment 2 (Rectangle, 4 to 6 s):
   $I_2 = \text{base} \times \text{height} = (6 - 4)\text{ s} \times 10\text{ N} = +20\text{ N}\cdot\text{s}$

   Segment 3 (Rectangle below axis, 6 to 8 s):
   $I_3 = \text{base} \times \text{height} = (8 - 6)\text{ s} \times (-5\text{ N}) = -10\text{ N}\cdot\text{s}$

   2. Determine Net Impulse

   $$I_{\text{net}} = I_1 + I_2 + I_3 = 20 + 20 - 10 = +30\text{ N}\cdot\text{s}$$

   3. Find the Final Velocity

   Using the impulse-momentum theorem:

   $$I_{\text{net}} = m(v_f - v_i) \implies 30 = 2(v_f - 3)$$

   $$15 = v_f - 3 \implies v_f = +18\text{ m/s}$$

   The force–time graph provides a direct method for finding velocity changes without needing to compute instantaneous acceleration at every point.