物理計算兩根繩索中的張力及最大加速度

(a) Since both the fireman (消防員) and parcel (包裹) are in equilibrium (平衡), net force (淨力) acting on them are zero.
Tension in the string holding the parcel = 10g N = 100 N
where g is the acceleration due to gravity(重力加速度) , taken to be 10 m/s^2

Tension in the string holding the fireman = (10 + 65)g N = 750 N

(b) Given the maximum horizontal component from the string = 900 N
hence, max tension in the string = 900/cos(30) N = 1039 N

Net force on the fireman and parcel = (1039 - 65g - 10g) N = 289 N
Apply: net force = mass x acceleration
289 = (65 + 10)a where a is the acceleration
hence, a = 289/75 m/s^2 = 3.85 m/s^2

圖:photo.pchome.com.tw/forever2015/139565640772