Forces and Free-Body Diagrams: Practice Questions (Physics 1 2.2)

Answer: (A). Once the beanbag leaves the hand, the only object interacting with it is Earth, through gravity. It keeps rising because it is already moving upward, not because a force pushes it up.

• (B) and (C) include a “force of the throw”. The hand stopped touching the beanbag when it let go, so no such force exists.
• (D) invents a normal force from the air. Normal forces come from a surface the object is pressed against. Air resistance is a separate force, and the question says it is negligible.

Answer: (A). Earth pulls straight down. The ramp pushes perpendicular to its surface. Friction acts along the surface, opposite to the block’s sliding, so up the slope.

• (B) draws only a component of gravity. Gravity is always vertical; “into the ramp” is just one of its components, and components are not drawn on the diagram.
• (C) adds a force with no object exerting it. The down-slope pull is already part of gravity (its mg sin θ component), so a separate arrow counts it twice.
• (D) makes the normal force vertical. That is only true on level ground; a surface pushes perpendicular to itself.

Answer: (A). Earth acts on the whole system. The snow touches the system (at the sled’s runners) and gives a normal force and friction. The rope is attached to the sled and pulls on it.

• (B) adds an internal force. The child and the sled are both inside the system, so the forces between them are not drawn.
• (C) adds a force on the rope, not on the system. The hand touches the rope, not the child or the sled.
• (D) leaves out friction and adds a force the system exerts on the snow. That force acts on the snow, so it belongs on the snow’s diagram.

(a) Three forces from one dot: F_g (Earth on crate) straight down; F_N (ramp on crate) perpendicular to the ramp, away from the surface; F_T (rope on crate) parallel to the ramp, up the slope. No friction arrow: the ramp is smooth. Axes: +x up the slope, +y perpendicular to the surface, away from it.

(b) F_g = (15 kg)(9.8 N/kg) = 147 N. Down the slope (−x): mg sin 20° = 147 × 0.342 = 50.3 N. Into the surface (−y): mg cos 20° = 147 × 0.940 = 138 N.

(c) x: F_T − 50.3 N = 0, so F_T = 50.3 N (about 50 N). y: F_N − 138 N = 0, so F_N = 138 N.

Suggested mark points (5): 1 for a diagram with exactly three correctly directed, labelled arrows and no friction; 1 for axes along and perpendicular to the ramp; 1 for mg sin 20° along the ramp; 1 for mg cos 20° perpendicular; 1 for F_T ≈ 50 N and F_N ≈ 138 N.

Common error: swapping sine and cosine gives 138 N down the slope. Check with a 0° ramp: there should be no down-slope component, and sin 0° = 0.

(a) Four forces from one dot (take the board to the eraser’s left):

• F_g, Earth on eraser: down, short;
• F_f, board on eraser: up, the same short length as F_g;
• F_A, finger on eraser: towards the board (left), long;
• F_N, board on eraser: away from the board (right), the same long length as F_A.

(b) Long-range: gravity only (Earth pulls on the eraser without touching it). Contact: the finger’s push, the normal force and friction. These are the large-scale result of electric forces between atoms at the touching surfaces: the finger presses the eraser against the board, the surface atoms are squeezed together, and their repulsion pushes back.

(c) Vertical: F_f = F_g = (0.040 kg)(9.8 N/kg) = 0.39 N, upward. Horizontal: F_N = F_A = 3.0 N, away from the board.

Suggested mark points (5): 1 for all four forces with correct directions; 1 for relative lengths (F_N = F_A, both much longer than F_g = F_f); 1 for correct classification; 1 for the electric origin of contact forces; 1 for both values in (c).

Common error: drawing the friction force downward or leaving it out. Without upward friction from the board, nothing would hold the eraser up against gravity.

Take forward as the direction of the child’s pull.

(a) Rear sled, four forces: F_g, Earth on rear sled (58.8 N, down); F_N, snow on rear sled (up); F_f, snow on rear sled (backward); F_T, joining rope on rear sled (forward).

(b) Front sled, five forces: F_g, Earth on front sled (78.4 N, down); F_N, snow on front sled (up); F_f, snow on front sled (backward); the child’s rope on the front sled (forward); the joining rope on the front sled (backward). The two backward forces are drawn side by side, not on top of each other.

(c) System, four forces: F_g, Earth on system (137 N, down); the snow’s normal force on the system (up); the snow’s friction on the system (backward); the child’s rope on the system (forward). The snow’s normal forces on the two sleds may be drawn as two side-by-side arrows or combined into one; the same goes for the two friction forces.

(d) The claim is incorrect. The joining rope is inside the chosen system. Its pull forward on the rear sled and its pull backward on the front sled are internal interactions between parts of the system. A free-body diagram shows only forces exerted on the system by its environment, so the joining rope’s tension does not appear. It would appear if the system were one sled only.

(a) No. Take the system to be the astronaut with her helmet and suit. Her hands and her helmet are both parts of this system, so the push is an internal interaction. An object or system cannot exert a net force on itself: as her hands push the helmet, the helmet pushes back on her hands, and the system as a whole gets no net force. Nothing outside the system touches her, so her center of mass stays at rest relative to the module.

(b) She can throw the tool away from the wall she wants to reach. While she pushes the tool, the tool pushes back on her. Now the tool is a separate object outside her system, so its push on her is an external force, and she starts to move towards the wall. Pushing the helmet failed because the helmet was part of her own system.

An answer that says “there is no gravity in space, so she cannot move” earns no credit: gravity is not what is missing. What is missing is an interaction with something outside her system.