Newton's Second Law in Two Dimensions
Apply F = ma along two perpendicular axes. For motion on a fixed straight slope, resolve forces along and perpendicular to the surface. While contact remains, perpendicular acceleration is zero.
What a learner can do afterwards
- Find the acceleration of a block sliding down a smooth slope
- Resolve perpendicular to the slope to find the normal reaction
- Handle a pulling force applied at an angle to the direction of travel
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You have used F = ma for resultant force and resolved forces into components. Apply F = ma along each perpendicular axis. For a fixed straight slope, use axes along and perpendicular to it. Let theta be its angle above horizontal, with 0 < theta < 90 degrees. Weight mg acts vertically down. Its components are mg sin(theta) downhill and mg cos(theta) into the slope. The normal reaction R points outward, perpendicular to the slope.
Smooth means no friction. If only weight and R act, mg sin(theta) = ma downhill, so a = g sin(theta). Mass cancels. With uphill positive, write -mg sin(theta) = ma. Acceleration is downhill even when velocity is uphill; the block then slows down. For a given g, increasing theta from 0 toward 90 degrees increases sin(theta) and downhill acceleration. Without extra forces, this acceleration is less than g.
While a block follows a fixed straight slope, perpendicular acceleration is zero. If only weight and R act perpendicular to it, R = mg cos(theta). Thus R < mg for 0 < theta < 90 degrees, but R = mg on level ground. To find mass, rearrange: m = R / (g cos(theta)).
With a parallel uphill pull P, choose uphill positive: P - mg sin(theta) = ma. For m = 6 kg, theta = 30 degrees, g = 10 m/s squared and P = 42 N, the resultant is 42 - 30 = 12 N uphill. With an extra parallel downhill force D, use downhill positive: D + mg sin(theta) = ma. Extra forces can make acceleration exceed g.
On a smooth level floor, a rope force F at angle alpha above horizontal has components F cos(alpha) forward and F sin(alpha) upward. If only rope, weight and reaction act, contact gives a = F cos(alpha)/m and R = mg - F sin(alpha). A 16 N pull at 60 degrees gives 8 N forward. At fixed F and m, increasing alpha from 0 toward 90 degrees reduces forward acceleration.
For a rope above a slope, measure alpha above the uphill direction. Then a = (F cos(alpha) - mg sin(theta))/m and R = mg cos(theta) - F sin(alpha), with uphill positive. An extra force Q perpendicular into a slope instead gives R = mg cos(theta) + Q. On level ground, the lifting threshold is F sin(alpha) = mg, where R = 0. A negative required R means contact cannot persist: the surface can push, but cannot pull the body toward itself.
Apply F = ma to the resultant force along each axis. On a fixed straight surface, balance perpendicular forces while contact remains, and include every applied force.
Lesson, questions, worked solutions and flashcards written for this dataset.
Readings, questions, worked solutions and flashcards written for this dataset. Linked videos, Newton's second law calculations and Forces at an angle, are credited to Khan Academy. No video license grant is asserted.
More to read
Two equations for one block
An 8 kg block slides on a fixed smooth slope at 45 degrees above horizontal. Only its weight and the surface reaction act. Take g = 10 m/s squared.
Its weight is 80 N vertically downward. Resolve this one force along two axes. The downhill component is 80 sin45 degrees, about 56.57 N. The inward perpendicular component is 80 cos45 degrees, also about 56.57 N. Do not add these components to the full weight again: they replace it in the component equations.
Along the slope, 80 sin45 degrees = 8a. Dividing by 8 gives a = 10 sin45 degrees, about 7.07 m/s squared downhill.
Perpendicular to the slope, the block has zero acceleration while it stays in contact. The reaction therefore balances the inward weight component: R = 80 cos45 degrees, about 56.57 N.
The block is moving and accelerating along the surface even though its perpendicular resultant is zero. Balance applies to one direction here, not to the whole force system.
When the floor stops pushing
A rope pulls a 5 kg body across a smooth level floor at 30 degrees above horizontal. Its force is 20 N. Only the rope, weight and floor reaction act. Take g = 10 m/s squared.
The body's weight is 50 N. The rope pulls upward with 20 sin30 degrees = 10 N. While contact remains, the vertical equation is R + 10 = 50, giving R = 40 N. Horizontally, the rope supplies 20 cos30 degrees, about 17.32 N. Dividing by 5 gives a forward acceleration of about 3.46 m/s squared.
Keep the rope angle fixed and increase its force. The upward component increases, so the required floor reaction decreases. At F sin30 degrees = 50 N, the required reaction reaches zero. This occurs at F = 100 N.
For a stronger pull, keeping the body on the floor would require a downward floor force. The floor cannot provide that force. The body lifts, and the zero-vertical-acceleration equation no longer describes its motion.
Moving uphill while accelerating downhill
A 6 kg block is moving uphill on a fixed smooth 30 degree slope. A rope pulls parallel uphill with 18 N. Only the rope, weight and normal reaction act. Take g = 10 m/s squared.
The downhill weight component is 6 × 10 × sin30 degrees = 30 N. Choose uphill as positive. The resultant along the slope is 18 - 30 = -12 N. Newton's law gives 6a = -12, so a = -2 m/s squared.
The minus sign describes downhill acceleration. It does not say that the block is already moving downhill. While it moves uphill, that downhill acceleration reduces its uphill velocity. If these forces continue to act and the slope is long enough, it reaches zero velocity and then starts moving downhill.
The direction of velocity comes from the motion at that instant. The direction of acceleration comes from the resultant force. Write the force equation before deciding the acceleration's sign. A normal reaction can be present throughout this motion without balancing the downhill force component.
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