The box now rests on a frictionless ramp angled at 15◦ . the mover pulls up on a rope attached to the box to move it up the incline. 33 kg f 29 ◦ 15◦ if the rope makes an angle of 29 ◦ with the horizontal, what is the smallest force the mover would have to exert to move the box up the ramp? the acceleration due to gravity is 9.81 m/s 2 .

F = 84.61 N

Explanation:

As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.

component of Force along the incline = F cos(23° - 15°) = F cos(8°)

component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)

Equating the above two components of forces will give the minimum Force required.

F cos(8°) = 33*9.81*sin(15°)

F = 33*9.81*sin(15°) / cos(8°)                  (calculate the value using a scientific calculator)

F = 84.61 N

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student