Question

worksheet: 2.1 forces and free body diagrams in 2d
● for each situation, sketch a system diagram and draw a free body diagram for the object bolded.
● for any object in equilibrium (experiencing uniform motion), determine the magnitude of the forces acting on the object.

1. a 5.0 kg mass falling in air at terminal velocity.
2. a 5.0 kg mass is pushed with uniform motion along the desk with 30n.
3. a 5.0 kg book that is held against a wall with a 30 n horizontal force.
4. a 5.0 kg sign hangs from two wires attached to the ceiling at an angle of 25° below the horizon.
5. a 5.0 kg mass on a ramp inclined at 25° to the horizon moving at:

a. 0 m/s
b. 1 m/s²

Explanation:

Step1: Analyze situation 1

At terminal velocity, net - force is zero. The force of gravity $F_g = mg$ and the air - resistance $F_{air}$ are equal. Given $m = 5.0\ kg$ and $g=9.8\ m/s^{2}$, $F_g=mg = 5.0\times9.8 = 49\ N$. So $F_{air}=49\ N$.

Step2: Analyze situation 2

Since the mass is in uniform motion, the net - force is zero. The applied force $F_{app}=30\ N$ and the frictional force $F_f$ are equal. So $F_f = 30\ N$. The normal force $F_N=mg=5.0\times9.8 = 49\ N$.

Step3: Analyze situation 3

In the horizontal direction, the normal force $F_N$ exerted by the wall on the book equals the applied horizontal force. So $F_N = 30\ N$. In the vertical direction, the force of gravity $F_g=mg = 5.0\times9.8=49\ N$ and the frictional force $F_f$ (holding the book from falling) are equal. So $F_f = 49\ N$.

Step4: Analyze situation 4

Let the tension in each wire be $T$. In the vertical direction, $2T\sin(25^{\circ})=mg$. Given $m = 5.0\ kg$ and $g = 9.8\ m/s^{2}$, we can solve for $T$: $T=\frac{mg}{2\sin(25^{\circ})}=\frac{5.0\times9.8}{2\times0.4226}\approx58\ N$.

Step5: Analyze situation 5a

When the mass is at rest ($a = 0\ m/s^{2}$), in the direction parallel to the ramp, $F_{parallel}=mg\sin(25^{\circ})=5.0\times9.8\times0.4226\approx21\ N$. The normal force $F_N=mg\cos(25^{\circ})=5.0\times9.8\times0.9063\approx44\ N$.

Step6: Analyze situation 5b

Using Newton's second law $F_{net}=ma$. In the direction parallel to the ramp, $F_{net}=mg\sin(25^{\circ})-F_f=ma$. Given $m = 5.0\ kg$, $a = 1\ m/s^{2}$, and $mg\sin(25^{\circ})\approx21\ N$. We can solve for the frictional force $F_f=mg\sin(25^{\circ})-ma=21 - 5.0\times1=16\ N$. The normal force $F_N=mg\cos(25^{\circ})\approx44\ N$.

1. System diagram: A dot representing the mass with an arrow downwards for gravity and an arrow upwards for air - resistance. Free - body diagram: Same as system diagram. Forces: $F_{air}=49\ N$, $F_g = 49\ N$.
2. System diagram: A rectangle representing the mass on a horizontal line (desk), with an arrow for the applied force, an arrow for the frictional force, an arrow downwards for gravity and an arrow upwards for the normal force. Free - body diagram: Same as system diagram. Forces: $F_{app}=30\ N$, $F_f = 30\ N$, $F_N=49\ N$, $F_g=49\ N$.
3. System diagram: A rectangle (book) against a vertical line (wall), with a horizontal arrow for the applied force, a horizontal arrow in the opposite direction for the normal force, a vertical arrow downwards for gravity and a vertical arrow upwards for the frictional force. Free - body diagram: Same as system diagram. Forces: $F_N = 30\ N$, $F_f=49\ N$, $F_g=49\ N$.
4. System diagram: A dot (sign) with two diagonal lines (wires) attached to a horizontal line (ceiling) at an angle of $25^{\circ}$ below the horizon, with a vertical arrow downwards for gravity. Free - body diagram: Same as system diagram. Forces: $T\approx58\ N$ (in each wire), $F_g=49\ N$.

5a. System diagram: A rectangle (mass) on a slanted line (ramp) at an angle of $25^{\circ}$ to the horizon, with an arrow parallel to the ramp (for the component of gravity along the ramp), an arrow perpendicular to the ramp for the normal force and an arrow downwards for gravity. Free - body diagram: Same as system diagram. Forces: $F_{parallel}\approx21\ N$, $F_N\approx44\ N$, $F_g=49\ N$.
5b. System diagram: Similar to 5a, but with an additional arrow for the frictional force. Free - body diagram: Same as system diagram. Forces: $F_f = 16\ N$, $F_N\approx44\ N$, $F_g=49\ N$, $F_{parallel}\approx21\ N$.

Answer:

1. System and free - body diagrams as described above. Forces: $F_{air}=49\ N$, $F_g = 49\ N$.
2. System and free - body diagrams as described above. Forces: $F_{app}=30\ N$, $F_f = 30\ N$, $F_N=49\ N$, $F_g=49\ N$.
3. System and free - body diagrams as described above. Forces: $F_N = 30\ N$, $F_f=49\ N$, $F_g=49\ N$.
4. System and free - body diagrams as described above. Forces: $T\approx58\ N$ (in each wire), $F_g=49\ N$.

5a. System and free - body diagrams as described above. Forces: $F_{parallel}\approx21\ N$, $F_N\approx44\ N$, $F_g=49\ N$.
5b. System and free - body diagrams as described above. Forces: $F_f = 16\ N$, $F_N\approx44\ N$, $F_g=49\ N$, $F_{parallel}\approx21\ N$.