Forces

A contact force acts only when objects touch. Friction acts along the surface between two touching objects. Gravity, magnetic and electric forces are field forces that act at a distance, and weight is just the name for the gravitational force.

Weight is the gravitational force: $W = mg = (5.0)(9.8) = 49\text{ N}$. (Weight is a force in newtons, not a mass in kilograms.)

Mass is the amount of matter and does not change with location. Weight ($W = mg$) depends on the local $g$, so it is smaller on the Moon even though the mass is unchanged.

On level ground with no vertical motion, the normal force balances the weight: $N = mg = (10)(9.8) = 98\text{ N}$.

Take right as positive: $\sum F = 12 - 5 = 7\text{ N}$ to the right. The vertical forces (weight and normal) cancel, so they do not affect the horizontal net force.

Constant velocity means no change in motion, so the object is in equilibrium and the net force is zero. The driving force is exactly balanced by friction and air resistance. (A moving object does not need a net force to keep moving.)

By Newton's third law, the swimmer pushes the water backward and the water pushes the swimmer forward with an equal and opposite force. It is this reaction force from the water, acting on the swimmer that drives them forward.

The forces are perpendicular, so combine them with Pythagoras: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5.0\text{ N}$. (Perpendicular forces are not simply added to $7\text{ N}$.)

The mass is in equilibrium, so the tension balances the weight: $T = mg = (6.0)(9.8) = 59\text{ N}$.

The normal force balances the part of the weight pressing into the surface: $N = mg\cos\theta = (4.0)(9.8)(0.87) \approx 34\text{ N}$. (The full weight is $39\text{ N}$, but only the perpendicular component presses on the slope.)

The component of weight along the slope is $mg\sin\theta = (4.0)(9.8)(0.5) = 19.6 \approx 20\text{ N}$. This is the force that would pull the block down a frictionless slope.

The upper rope supports both masses, a total of $5.0\text{ kg}$: $T = (3.0 + 2.0)(9.8) = 49\text{ N}$. (The lower rope carries only the $2.0\text{ kg}$ mass, so its tension is $20\text{ N}$.)

Only the horizontal part of the pull drives the box: $30\cos 40^\circ = 30 \times 0.77 = 23.1\text{ N}$. Subtract friction: $\sum F = 23.1 - 10 \approx 13\text{ N}$. (Forgetting to resolve gives the $20\text{ N}$ trap.)

A third-law pair acts on two different objects and is the same type of force. The Earth pulls the book down (its weight), so the reaction is the book pulling the Earth up. The normal force is a separate contact force, and it equals the weight here only because the book is in equilibrium, not because they are an action-reaction pair.

By Newton's third law the two forces are equal in size and opposite in direction. The bullet accelerates far more only because its mass is much smaller ($a = F/m$), not because the force on it is larger.

Along the slope the tension balances the component of weight pulling the block down: $T = mg\sin\theta = (5.0)(9.8)(0.5) = 24.5\text{ N}$. (The full weight is $49\text{ N}$, but only the along-slope part must be held.)

Pushing partly downward makes the box press harder on the floor. Balancing the vertical forces: $N = mg + F\sin\theta = (5.0)(9.8) + (20)(0.5) = 49 + 10 = 59\text{ N}$. The normal force is not always equal to $mg$ — pulling up at an angle would instead reduce it below $49\text{ N}$.

The two vertical components share the weight: $2T\sin 60^\circ = mg$, so $T = \dfrac{mg}{2\sin 60^\circ} = \dfrac{(20)(9.8)}{2(0.87)} \approx 113\text{ N}$. Each rope carries more than half the weight because it also pulls sideways.

The two forces have a resultant of $\sqrt{6^2 + 8^2} = 10\text{ N}$ pointing north-east. The equilibrant is equal and opposite: $10\text{ N}$ pointing south-west, so the three forces sum to zero.

Only the ceiling rope has a vertical component, so it alone supports the weight: $T\cos 30^\circ = mg$, giving $T = \dfrac{mg}{\cos 30^\circ} = \dfrac{(8.0)(9.8)}{0.87} \approx 90\text{ N}$. (The horizontal rope only pulls sideways.)

The action-reaction pair (horse on cart, cart on horse) acts on different objects, so it never cancels for the system. To see the system's motion, look at the external forces: the ground pushes the horse forward (reaction to its hooves pushing back) more than friction and drag resist the cart, so the whole system accelerates.

The up-slope direction is $30^\circ$ above the horizontal, so the horizontal force resolves along the slope as $F\cos 30^\circ = 12 \times 0.87 = 10.4\text{ N}$. (Using $\sin$ instead gives the $6\text{ N}$ trap.)

Only the angled rope has a vertical component to hold up the weight: $T\sin 40^\circ = mg$, so $T = \dfrac{mg}{\sin 40^\circ} = \dfrac{(15)(9.8)}{0.64} \approx 230\text{ N}$. (Forgetting to divide by $\sin 40^\circ$ gives the $147\text{ N}$ trap, which is just the weight.)

Resolving along the slope, the up-slope part of $F$ must balance the down-slope part of the weight: $F\cos 40^\circ = mg\sin 40^\circ$, so $F = mg\tan 40^\circ = (6.0)(9.8)(0.84) \approx 49\text{ N}$.

The force is smallest when it points straight up the slope, so all of it opposes the down-slope weight component: $F = mg\sin\theta = (8.0)(9.8)(0.5) = 39\text{ N}$. A horizontal or any other direction wastes part of the force pressing into the slope, so a larger force would be needed.