Question

The force table consists of a circular platform with marking at the edge showing the angle for a circle centered about the center of the table. A mass, m = 120 gram is places on a hanger which has a mass of 50 g, giving a total mass of 170 g. The hanger is hung at an angle of 110 deg over a pulley attached to the force table using a string of negligible mass. The pulley over which the string rests has very low friction force.

The tension in the string: 1.67  N.
The components: Fx = -0.570 N, Fy = 1.57 N
Total mass needed to balance the force: 170 g
Angle should the mass be hung: 290 deg.

This is where I'm confused and having problems:
The balancing mass used in the last exercise is kept in place. The original mass that was hanging at 110 deg is removed and replaced with its components with the x and y axes at 0 deg and the 90deg respectively. What is the total mass (mass plus hanger) needed along the x-axis? Along which direction should the mass be hung?
Mass = ??
I know that it will be -x from the Fx component

What is the total mass (mass plus hanger) needed along the y-axis? Along which direction should the mass be hung?
Mass = ??
I know that it will be +y from the Fy component

Answer 1

The force table consists of a circular platform with marking at the edge showing the angle for a circle centered about the center of the table. A mass, m = 220 gram is places on a hanger which has a mass of 50 g, giving a total mass of 270 g. The hanger is hung at an angle of 140 deg over a pulley attached to the force table using a string of negligible mass. The pulley over which the string rests has very low friction force.

The tension in the string: 265 N.
The components: Fx = -2.03 N, Fy = 1.70 N
Total mass needed to balance the force: 270 g
Angle should the mass be hung: 320 deg.

The horizontal component of the 270g weight (270g=0.27kg, 50g = 0.05kg)
0.27*9.8*cos140 = -2.03N (negative means towards 180deg) = 9.8*(0.05+mx)
=> mx = 156.8g 156.8+50=206.8g including hanger
Hung at 0 degrees to counteract horizontal force component at 180 degrees
The vertical component of the 270g weight
0.27*9.8*sin140 = 1.70N = 9.8*(0.05+my) => my = 123.6g
Postive meaning towards 90 degrees
123.6+50=173.6g including hanger
Hung at 270 degrees