Questions: QUESTION 11
Use the following to answer questions 11 thru 14. Two blocks of masses m1=12.0 kg and m2=30.0 kg are connected by a light string as shown below. The coefficient of kinetic friction between m1 and the table is μ=0.250 and m2 is initially located at y=6.0 meters above the floor. Assuming the system starts from rest, draw a free body diagram for each block and determine the normal force acting upon m1.
Transcript text: QUESTION 11
Use the following to answer questions 11 thru 14.
Two blocks of masses $\mathrm{m}_{1}=12.0 \mathrm{~kg}$ and $\mathrm{m}_{2}=30.0 \mathrm{~kg}$ are connected by a light string as shown below. The coefficient of kinetic friction between $\mathrm{m}_{1}$ and the table is $\mu=0.250$ and $\mathrm{m}_{2}$ is initially located at $\mathrm{y}=6.0$ meters above the floor. Assuming the system starts from rest, draw a free body diagram for each block and determine the normal force acting upon $\mathrm{m}_{1}$.
Solution
Solution Steps
Step 1: Analyzing the forces acting on m1
The forces acting on m1 are:
Weight (W1): Downward force due to gravity, equal to m1 * g.
Normal force (N1): Upward force exerted by the table on m1, perpendicular to the surface.
Tension (T): Force exerted by the string, pulling m1 horizontally to the right.
Friction force (f): Force opposing the motion of m1, acting horizontally to the left.
Step 2: Determining the normal force on m1
Since m1 is not accelerating vertically, the net vertical force is zero. This means the normal force must balance the weight of m1. Therefore, N1 = W1 = m1 * g.
Step 3: Calculating the normal force
Given m1 = 12.0 kg and g = 9.8 m/s² (approximate value for acceleration due to gravity), we can calculate N1: