When a vehicle accelerates with acceleration a, a bob hanging from its roof tilts backward (opposite to the motion) and settles at an angle θ from the vertical given by tan θ = a/g. Inside the vehicle a pseudo-force ma pushes the bob backward, while gravity mg pulls it down, and the string lines up along the sum of these two. Memory hook: "The bob leans back, and tan of the lean equals a over g" — the bob is lazy, so it lags behind the vehicle that speeds up under it.
Left: the bob hangs backward at angle θ from the vertical while the vehicle accelerates right. Right: free-body diagram in the vehicle frame — weight mg down, pseudo-force ma backward, tension T along the string; their balance gives tan θ = a/g.
Your doubts, answered
Which direction does the bob tilt — forward or backward?
The bob tilts backward, opposite to the direction the vehicle accelerates. The bob has inertia and tends to stay at rest, so when the vehicle jumps forward the bob lags behind and swings back. In the vehicle frame we say a pseudo-force ma acts on the bob in the backward direction, which is why the string leans backward.
Is the angle tan⁻¹(a/g) or sin⁻¹(a/g)?
It is tan⁻¹(a/g). The horizontal pseudo-force is ma and the vertical weight is mg. These are the two perpendicular sides of a right triangle, so their ratio gives the tangent: tan θ = ma/mg = a/g. Using sin⁻¹(a/g) is a common trap and is wrong because a and g are not the two forces on the same line; they are the two perpendicular components.
What is the tension in the string here?
The string must balance both the weight mg (vertical) and the pseudo-force ma (horizontal). So tension T = √((ma)² + (mg)²) = m√(a² + g²). This is always more than mg, because the string now supports an extra sideways pull as well as the weight.
Does a heavier bob tilt more?
No. The angle does not depend on mass at all. tan θ = ma/mg, and the mass m cancels from top and bottom, giving tan θ = a/g. A light bob and a heavy bob hanging in the same vehicle tilt by exactly the same angle.
Why can I treat this like a stationary equilibrium problem?
Because we work in the vehicle's frame, which is non-inertial (it is accelerating). To use equilibrium there, we add the pseudo-force ma on the bob pointing backward. Now the bob is at rest relative to the vehicle, so the three forces — weight, tension, pseudo-force — are in balance and we can apply the equilibrium conditions ΣFx = 0 and ΣFy = 0.
⚠️ The NEET trap ✗ θ = sin⁻¹(a/g), tilting forward in the direction of motion. ✓ θ = tan⁻¹(a/g), tilting backward, opposite to the acceleration. 🧠 a and g are the two perpendicular sides of the force triangle, so their ratio is a TANGENT, not a sine. And the bob lags BACK, never leaning forward into the motion.
Real NEET questions
NEET 2024
A bob is suspended by a light string from the roof of a truck. The truck, initially stationary, suddenly moves to the right with acceleration a. The pendulum will tilt:
A · to the left, with inclination sin⁻¹(a/g) to the vertical
B · to the left, with inclination tan⁻¹(a/g) to the vertical ✓
C · to the right, with inclination sin⁻¹(a/g) to the vertical
D · to the left, with inclination tan⁻¹(g/a) to the vertical
Solution: Work in the truck's frame. The truck accelerates to the right with a, so a pseudo-force of magnitude ma acts on the bob toward the LEFT (opposite to the acceleration). Gravity mg acts downward. Set up equilibrium of the bob relative to the truck. Horizontal: T sin θ = ma. Vertical: T cos θ = mg. Divide the two equations: (T sin θ)/(T cos θ) = ma/mg, so tan θ = a/g, giving θ = tan⁻¹(a/g). The horizontal push is toward the left, so the bob tilts to the left (backward). Answer: to the left, with inclination tan⁻¹(a/g) — option B.
Solved Laws Of Motion NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for the angle of a bob in an accelerating vehicle?
tan θ = a/g, so θ = tan⁻¹(a/g), where a is the vehicle's acceleration and g is gravity. The angle is measured from the vertical.
Why does the bob tilt backward and not forward?
Because of inertia. The bob tends to stay where it is while the vehicle speeds forward under it, so the bob lags behind. In the vehicle frame this is described by a backward pseudo-force ma.
Does this angle depend on the length of the string or the mass of the bob?
No. Neither length nor mass appears in tan θ = a/g. Only the vehicle's acceleration a and gravity g decide the tilt angle.
What is the tension in the string?
T = m√(a² + g²). It is the resultant needed to balance both the weight mg and the pseudo-force ma, so it is always larger than mg.
How is this useful for NEET?
This is a direct, high-yield pseudo-force result that NEET has asked (e.g. 2024). Remembering tan θ = a/g, the backward tilt, and T = m√(a² + g²) lets you solve these questions in seconds without drawing long derivations.
