A 2.5kg box and 5kg box are placed on a sewsaw apparatus. They both rest at the

Question

A 2.5kg box and 5kg box are placed on a sewsaw apparatus. They both rest at the ends of the plank that is rigid with negliable mass. The plank is 2m in length and raised above the ground 20cm on the pivot point. Assume that the system is balanced and consider the boxes as point particles.

a) Find the pivot point from the 2.5kg box

b) Find the force on the pivot point

c) Find the angular acceleration when you place an additional 4kg box next to the 5kg box (assume the system is balanced right now).

d) Now assume that the seesaw tips to the heavier side, hitting the ground. However, the 2.5kg box doesn't slide down the plank so it must be staying there due to static friction. Find the coefficient of static friction holding the 2.5kg box in place.

e) Now if you slide the 4kg box (the one you added on) towards the center of the plank the seesaw will become balanced again. Find the point at which the system is balanced again.

Explanation / Answer

a)
Let the distance of pivot point be x m from 2.5 Kg box
use:
Torque due to 2.5 Kg = Torque due to 5 Kg
2.5*g*x = 5*g*(2-x)
2.5*x = 5*(2-x)
2.5x = 10 - 5x
7.5x = 10
x = 1.33 m
Answer: 1.33 m

b)
Force at pivot point must balance weight of 2.5 Kg and 5 Kg
F = (2.5 + 5)* g
F = (2.5 + 5)* 9.8
= 73.5 N
Answer: 73.5 N

c)
If the system is balanced, wont the angular acceleration be 0
Answer: 0

d)
let the tip angle be thetha
h = 20 cm = 0.2 m
d = 2 -1.33 = 0.67 m
thetha = asin (0.2/0.67)
= 17.4 degree

static friction must balance the compoenent of weight along the incline
static friction = compoenent of weight along incline
miu*m*g*cos thetha = m*g* sin thetha
miu*cos thetha = sin thetha
miu = tan thetha
= tan 17.4
= 0.313
Answer: 0.313

Only 4 subparts at a time please

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