An engineer designs a lift that takes passengers up a slope along a straight line. The floor of the lift is kept horizontal.
Suppose a man of mass 65 kg takes the lift up the slope. Find the normal force N and friction f (magnitude and direction )acting on the man by the floor of the lift when the lift accelerates at 0.5 m/s^(2)
I actually want the steps, please!!!
- 天同Lv 75 年前最愛解答
Resolve the acceleration into vertical and horizontal directions.
Vertical acceleration = 0.5.sin(20) m/s^2 = 0.171 m/s^2
Horizontal acceleration = 0.5.cos(20) m/s^2 = 0.4698 m/s^2
Consider vertical motion, use: net-force = mass x acceleration
N - 65g = 65 x 0.171 where N is the normal reaction on the man, and g is the acceleration due to gravity (= 9.81 m/s^2)
i.e. N = (65 x 0.171 + 65g) N = 653 N
Consider the horizontal motion, use again: net-force = mass x acceleration
f = 65 x 0.4698 where f is the frictional force acting on the man
i.e. f = 30.5 N
The frictional force is acting horizontally in the forward direction, i.e. towards the right.
- 匿名5 年前
Take upwards and the direction to the right as positive
consider the vertical direction.
N-65 x 9.81 = 65 x 0.5 sin20
The normal reaction is 649N upwards.
Consider the horizontal direction.
f=65 x 0.5 cos20
The friction is 30.5 N towards the right.