Question

1. draw a free body diagram depicting the various forces at play in these static systems: a. mass & pulley system on a frictionless inclined plane: b. gondola lift: c. mass on an inclined plane (with friction):

Explanation:

Step1: Analyze mass - pulley system on frictionless inclined plane

• The mass on the inclined plane has gravitational force $mg$ acting down - ward. Resolve it into components: $mg\sin\theta$ along the incline and $mg\cos\theta$ perpendicular to the incline. The normal force $N = mg\cos\theta$ acts perpendicular to the surface of the incline. The tension $T$ in the string acts along the string, pulling the mass up the incline.

Step2: Analyze gondola lift

• The gondola has a weight $mg$ acting downward. There are two tension forces $T_1$ and $T_2$ in the ropes on either side of the gondola, making angles $\theta_1$ and $\theta_2$ with the horizontal.

Step3: Analyze mass on inclined plane with friction

• The mass has a gravitational force $mg$ acting downward. Resolve it into $mg\sin\theta$ along the incline and $mg\cos\theta$ perpendicular to the incline. The normal force $N = mg\cos\theta$ acts perpendicular to the incline. The frictional force $f=\mu N=\mu mg\cos\theta$ acts along the incline, opposing the tendency of the mass to slide. The direction of the frictional force depends on the motion or impending motion of the mass. If the mass is on the verge of sliding down, $f$ acts up - the incline; if it is being pulled up, $f$ acts down - the incline.

Answer:

For the mass - pulley system on a frictionless inclined plane:

• Draw a block on the incline. Draw a vector for the gravitational force $mg$ downward, a normal force $N$ perpendicular to the incline, and a tension force $T$ along the string.

For the gondola lift:

• Draw the gondola. Draw a vector for the weight $mg$ downward and two tension vectors $T_1$ and $T_2$ at angles $\theta_1$ and $\theta_2$ with the horizontal.

For the mass on an inclined plane with friction:

• Draw a block on the incline. Draw a gravitational force $mg$ downward, a normal force $N$ perpendicular to the incline, and a frictional force $f$ along the incline (direction as per the analysis above) and resolve $mg$ into $mg\sin\theta$ along the incline and $mg\cos\theta$ perpendicular to the incline.