Static Friction on a Ramp with a Hanging Block
On a laboratory ramp inclined at 30.0° to the horizontal, a 4.0 kg crate is connected by a light string over a frictionless pulley to a hanging 2.5 kg block. The string is parallel to the ramp. The coefficients of friction between crate and ramp are μs = 0.45 and μk = 0.30. The system is released from rest; the ramp is 2.0 m long, and the hanging block is initially 1.5 m above the floor. Use g = 9.8 m/s². A student predicts that the 2.5 kg block will accelerate downward because its weight is greater than the component of the crate’s weight along the ramp. The lab instructor asks you to check the prediction before the apparatus is released, and then to consider what happens if the hanging block is replaced by a 1.0 kg block.
1. Carry out the static-friction check for the 2.5 kg hanging block. Does the system move? Show the numerical comparison explicitly.
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Take up the ramp as positive for the crate. The hanging block’s weight is W₂ = m₂g = 2.5 × 9.8 = 24.5 N. The crate’s weight component along the ramp is m₁g sin30° = 4.0 × 9.8 × 0.500 = 19.6 N down the ramp. With no friction the net driving force would be 24.5 − 19.6 = 4.9 N up the ramp.
The maximum static friction is f_s,max = μs m₁g cos30° = 0.45 × 4.0 × 9.8 × 0.866 = 15.3 N. Since 4.9 N < 15.3 N, static friction can hold the system at rest. The student’s prediction fails because it ignores static friction; the block does not accelerate.
2. For the 2.5 kg case at rest, find the tension and the static friction force on the crate, giving magnitude and direction.
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At rest the hanging block is in equilibrium, so T = m₂g = 24.5 N. For the crate, T acts up the ramp with 24.5 N and the gravity component acts down the ramp with 19.6 N. The required friction is therefore 24.5 − 19.6 = 4.9 N down the ramp. This is below the 15.3 N static limit, so the crate remains at rest.
The tension is not equal to the difference of the two weights; it is fixed by the hanging block’s equilibrium. Static friction adjusts to 4.9 N down the ramp to keep the crate from sliding upward.
3. Repeat the check for a 1.0 kg hanging block. Does the crate slide? What is the tension?
Read the worked answer
With m₂ = 1.0 kg, W₂ = 1.0 × 9.8 = 9.8 N. The crate’s gravity component is still 19.6 N down the ramp, so without friction there would be a net 19.6 − 9.8 = 9.8 N down the ramp. The maximum static friction is still 15.3 N. Since 9.8 N < 15.3 N, the crate does not slide. Friction is 9.8 N up the ramp.
The hanging block is at rest, so T = 9.8 N. The system remains static in both cases, but the direction of the required static friction has reversed.