# Physics Question?

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!!!

### 2 個解答

• 天同
Lv 7
5 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.

apply F=ma

N-65 x 9.81 = 65 x 0.5 sin20

N=649N

The normal reaction is 649N upwards.

Consider the horizontal direction.

f=65 x 0.5 cos20

f=30.5 N

The friction is 30.5 N towards the right.
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