1 A ball of mass m is suspended

• 1. A ball of mass m is suspended from two strings of unequal length as shown above. The magnitudes of the tensions T 1 and T 2 in the strings must satisfy which of the following relations? • (A) Tl = T 2 (B) T 1 > T 2 (C) T 1 < T 2 (D) Tl + T 2 = mg

Answer • As T 2 is more vertical, it is supporting more of the weight of the ball. The horizontal components of T 1 and T 2 are equal. • C

Answer • 2. Normal force is perpendicular to the incline, friction acts up, parallel to the incline (opposite the motion of the block), gravity acts straight down. • E

Which of the following correctly indicates the magnitudes of the forces acting up and down the incline? • (A) 20 N down the plane, 16 N up the plane • (B) 4 N down the plane, 4 N up the plane • (C) 0 N down the plane, 4 N up the plane • (D) 10 N down the plane, 6 N up the plane

Answer • The component of the weight down the plane is 20 N sin 30. The net force is 4 N, so the friction force up the plane must be 4 N less than 10 N. • D

A plane 5 meters in length is inclined at an angle of 37°, as shown above. A block of weight 20 N is placed at the top of the plane and allowed to slide down. The magnitude of the normal force exerted on the block by the plane is • (A) greater than 20 N • (B) greater than zero but less than 20 N • (C) equal to 20 N • (D) zero

Answer • The weight component perpendicular to the plane is 20 N sin 37 o. To get equilibrium perpendicular to the plane, the normal force must equal this weight component, which must be less than 20 N. • B

• a. • b. f = μN where N = m 1 g cos θ gives μ = f /(m 1 g cos θ ) • c. constant velocity means ΣF = 0 where Σ Fexternal = m 1 g sin θ + m 2 g sin θ – f – 2 f – Mg = 0 • solving for M gives M = (m 1 + m 2) sin θ – (3 f)/g • d. Applying Newton’s second law to block 1 gives Σ F = m 1 g sin θ – f = m 1 a which gives a = g sin θ – f/m 1