9. A PH2 11 lab that used to be done here involved a force table (as shown, appr
ID: 1375495 • Letter: 9
Question
9. A PH2 11 lab that used to be done here involved a force table (as shown, approximately.) Different forces could be applied to the strings at different angles. The strings all met at a center pin. The forces could be varied by adding or subtracting masses to the hangers and the angles could be varied by moving the pulleys to different positions around the edge of the table. The point of the lab was ?vector addition?. One activity was to keep the center pin centered, meaning all the forces are in equilibrium, meaning all the forces sum to zero. Force one: 2.45 N at 180.0 degree. Force two: 3.55 N at 120.0 degree. Force three: 2.95 N at 20.0 degree. If the pin in the center of the table is in equilibrium, what (and at what angle) is force four?Explanation / Answer
let fourth force be F and at angle t
then balancing the x component:
2.45*cos(180)+3.55*cos(120)+2.95*cos(20)+F*cos(t)=0
F*cos(t)=1.4529 ..(1)
2.45*sin(180)+3.55*sin(120)+2.95*sin(20)+F*sin(t)=0
F*sin(t)=-4.08 N
diviing 1 by 2,
t=-70.414 degrees or 289.59 degrees
F=4.3341 N
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