

Guide 52. 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 1dimensional 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 xaxis to the right in the direction of the acceleration. We choose the positive yaxis up. +x +y Step 3. List the givens and goal Given: T = 44 N Note that the pull
force is a tension force. g = 9.8 m/s^{2} is understood for locations on the surface of the Earth. Goals: Find the acceleration, a, of 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 ycomponents. The tension force has both x and ycomponents. T_{x} = TcosθT_{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 xequation for the acceleration, and we solve the yequation 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: 

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