2. 1. A ball of mass m is suspended from two strings of unequal length as shown above. The magnitudes of the tensions T1 and T2 in the strings must satisfy which of the following relations?
(A) Tl = T2 (B) T1 > T2 (C) T1 < T2 (D) Tl + T2 = mg

3. Answer
As T2 is more vertical, it is supporting more of the weight of the ball. The horizontal components of T1 and T2 are equal. C

5. 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

6. 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

7. 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

8. 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

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

11. a.
b. f = μN where N = m1g cos θ gives μ = f /(m1g cos θ )
c. constant velocity means ΣF = 0 where Σ Fexternal = m1g sin θ + m2g sin θ – f – 2f – Mg = 0
solving for M gives M = (m1 + m2) sin θ – (3f)/g
d. Applying Newton’s second law to block 1 gives Σ F = m1g sin θ – f = m1a which gives a = g sin θ – f/m1