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 whole - system for acceleration. The net force is the difference in gravitational forces of the hanging masses. The total mass of the system is $m_{total}=12 + 10=22$ kg. The net force $F_{net}=(12\times9.8)-(10\times9.8)=(12 - 10)\times9.8$ N. Using $F_{net}=m_{total}a$, we have $a=\frac{(12\times9.8 - 10\times9.8)}{12 + 10}=\frac{(12 - 10)\times9.8}{22}\approx0.89$ m/s². For tension, consider one of the masses, say the 10 - kg mass. $T - 10\times9.8=10\times a$. Substituting $a = 0.89$ m/s², we get $T=10\times(9.8 + 0.89)=107.8$ N.

Step2: Analyze system b

The total mass of the system is $m_{total}=10 + 12 = 22$ kg. The net force is the gravitational force of the 12 - kg hanging mass, $F_{net}=12\times9.8$ N. Using $F_{net}=m_{total}a$, we have $a=\frac{12\times9.8}{10 + 12}=\frac{12\times9.8}{22}\approx5.24$ m/s². For the 10 - kg mass on the table, $T=10\times a=10\times5.24 = 52.4$ N.

Step3: Analyze system c

The total mass of the system is $m_{total}=40+25 = 65$ kg. The frictional force on the 40 - kg mass is $F_f=\mu N=\mu\times40\times9.8$, where $\mu = 0.40$. The net force is $F_{net}=25\times9.8-\mu\times40\times9.8$. Using $F_{net}=m_{total}a$, we have $a=\frac{25\times9.8-0.4\times40\times9.8}{40 + 25}=\frac{(25-16)\times9.8}{65}=\frac{9\times9.8}{65}\approx1.35$ m/s². For tension, consider the 25 - kg mass. $25\times9.8 - T=25\times a$. Substituting $a\approx1.35$ m/s², we get $T = 25\times(9.8 - 1.35)=25\times8.45 = 211.25$ N.

Step4: Analyze system d

The total mass of the system is $m_{total}=5+15 = 20$ kg. The net force is the gravitational force of the 15 - kg hanging mass, $F_{net}=15\times9.8$ N. Using $F_{net}=m_{total}a$, we have $a=\frac{15\times9.8}{5 + 15}=\frac{15\times9.8}{20}=7.35$ m/s². For the 5 - kg mass on the table, $T=5\times a=5\times7.35 = 36.75$ N.

Step5: Analyze system e

The total mass of the system is $m_{total}=16 + 8+24=48$ kg. The frictional force on the 8 - kg mass is $F_f=\mu N=\mu\times8\times9.8$, where $\mu = 0.20$. The net force is $F_{net}=(24\times9.8-16\times9.8)-\mu\times8\times9.8$. Using $F_{net}=m_{total}a$, we have $a=\frac{(24 - 16)\times9.8-0.2\times8\times9.8}{16 + 8+24}=\frac{(8 - 1.6)\times9.8}{48}=\frac{6.4\times9.8}{48}\approx1.31$ m/s².

Answer:

System a: $a\approx0.89$ m/s², $T\approx107.8$ N
System b: $a\approx5.24$ m/s², $T = 52.4$ N
System c: $a\approx1.35$ m/s², $T\approx211.25$ N
System d: $a = 7.35$ m/s², $T = 36.75$ N
System e: $a\approx1.31$ m/s²