Solving a Net Force Problem in Two Dimensions

Guide 5-2. Solving Net Force Problems in Two Dimensions

Prerequisite: Study section 5.7.

The following example problem illustrates the method for solving a net force problem in two dimensions. The only difference from a 1-dimensional problem is that you have to resolve forces into components and write a net force equation for each axis.

Problem: A child pulls a wagon with a force of 44 N by a handle making an angle of 29° with the horizontal. If the wagon has a mass of 4.5 kg, what is the acceleration of the wagon? What is the force with which the ground pushes up on the wagon?

Step 1. Draw a picture

A picture is drawn showing the directions of the acceleration and velocity.

Step 2. Choose the directions of +x and +y

We choose the positive x-axis to the right in the direction of the acceleration. We choose the positive y-axis up.

Step 3. List the givens and goal

Given:

T = 44 N Note that the pull force is a tension force.
m = 4.5 kg
θ = 29°
a_y = 0 We state that the vertical acceleration is 0.

g = 9.8 m/s² is understood for locations on the surface of the Earth.

Goals:

Find the acceleration, a, of the wagon
Find the normal force, N, of the ground on the wagon

Step 4. Draw a force diagram

The wagon is represented as a dot. The +x and +y axes are shown on the diagram together with the three relevant forces.

Forces on the wagon

Step 5. Resolve the forces into components and write the net force equations

The normal and weight forces have only y-components. The tension force has both x- and y-components.

T_x = Tcosθ

T_y = Tsinθ

There is a net force equation for each coordinate axis.

F_net,x = T_x = Tcosθ

F_net,y = Tsinθ + N - mg

Step 6. Apply Newton's 2nd Law

We apply Newton's 2nd Law for each coordinate axis.

ma_x = Tcosθ

ma_y = Tsinθ + N - mg

Step 7. Solve algebraically

We solve the x-equation for the acceleration, and we solve the y-equation for the normal force.

a_x = Tcosθ/m

The vertical acceleration is 0, so

0 = Tsinθ + N - mg,

N = mg - Tsinθ

Step 8. Substitute and reduce

a_x = (44 N)cos29°/4.5 kg = 8.6 m/s²

N = (4.5 kg)(9.8 m/s²) - (44 N)sin29° = 23 N

Step 9. Check your answer

The units work out since N = kgm/s². Acceleration is positive which is to the right, as expected. The normal force is positive, since we're treating it as a magnitude. The value of the normal force is less than the weight, since the vertical component of the tension force works together with the normal force to balance the weight of the wagon. The acceleration is high but at least it's less than g.

Here are 2 additional examples: