Question

remember. acceleration is determined with the whole system (all masses) tension is determined with just one mass (use only 1 part of diagram). tension should be equal for the system (ie. t1 = t2 so they cancel each other out) redraw your fbd in a horizontal plane to make it easier. 1. determine the acceleration and tension in each system below. a. b. c. d. e. (only calculate acceleration for e)

Explanation:

Step1: Analyze system a

Consider the two - mass system with hanging and horizontal masses. The net force acting on the whole system is the difference in gravitational forces of the hanging masses. The total mass of the system $m_{total}=12 + 10+12+10=44$ kg. The net force $F_{net}=(12 - 10)\times9.8$ N. According to Newton's second law $F = ma$, so $a=\frac{(12 - 10)\times9.8}{44}\approx0.445$ m/s². To find the tension, consider one of the masses. Let's take the 12 - kg mass on the left. $F_{g1}-T_{1}=m_{1}a$, where $F_{g1}=12\times9.8$ N and $m_{1} = 12$ kg. Solving for $T_{1}$, we get $T_{1}=12\times(9.8 - 0.445)=112.26$ N.

Step2: Analyze system b

The total mass of the system $m_{total}=10 + 12=22$ kg. The net - force is due to the hanging 12 - kg mass, $F_{net}=12\times9.8$ N. Using $F = ma$, $a=\frac{12\times9.8}{22}\approx5.35$ m/s². For the tension, consider the 10 - kg mass on the table. $T_{1}=10\times a = 10\times5.35 = 53.5$ N.

Step3: Analyze system c

The normal force on the 40 - kg mass on the table $N = 40\times9.8$ N. The frictional force $F_f=\mu N=0.4\times40\times9.8$ N. The net force on the system is $F_{net}=25\times9.8-0.4\times40\times9.8$. The total mass of the system $m_{total}=40 + 25=65$ kg. Using $F = ma$, $a=\frac{25\times9.8-0.4\times40\times9.8}{65}=\frac{(25 - 16)\times9.8}{65}\approx1.32$ m/s². To find the tension, consider the 25 - kg mass. $25\times9.8-T = 25\times a$, so $T=25\times(9.8 - 1.32)=212$ N.

Step4: Analyze system d

The total mass of the system $m_{total}=15 + 5+10=30$ kg. The net force $F_{net}=(15 - 10)\times9.8$ N. Using $F = ma$, $a=\frac{(15 - 10)\times9.8}{30}\approx1.63$ m/s². To find the tension, consider the 10 - kg mass. $T_{2}-10\times9.8=10\times a$, $T_{2}=10\times(9.8 + 1.63)=114.3$ N.

Step5: Analyze system e

The normal force on the 80 - kg mass $N = 80\times9.8$ N. The frictional force $F_f=\mu N=0.2\times80\times9.8$ N. The net force on the system $F_{net}=(24 - 16)\times9.8-0.2\times80\times9.8$. The total mass of the system $m_{total}=80 + 16+24=120$ kg. Using $F = ma$, $a=\frac{(24 - 16)\times9.8-0.2\times80\times9.8}{120}=\frac{(8 - 16)\times9.8}{120}\approx - 0.65$ m/s² (the negative sign just indicates the direction is opposite to the assumed positive direction).

Answer:

a. Acceleration: $0.445$ m/s², Tension: $112.26$ N
b. Acceleration: $5.35$ m/s², Tension: $53.5$ N
c. Acceleration: $1.32$ m/s², Tension: $212$ N
d. Acceleration: $1.63$ m/s², Tension: $114.3$ N
e. Acceleration: $- 0.65$ m/s²