求作用於方塊上的摩擦力。

I suppose there should be a diagram on how the spring balance is hooked to the block. Without the diagram, I have to assume that the spring balance is hooked to the upper side of the block preventing it from sliding down the inclined plane.

Assume that the block is in equilibrium, we have,
g.sin(30) = T + f
i.e. f = g.sin(30) - T = (10.sin(30) - 3.5) N = 1.5 N
where g is the acceleration due to gravity, taken to be 10 m/s^2


2014-03-29 10:28:27 補充：
Q: 彈簧秤讀數為3.5N,3.5N指的不是(T + f)嗎?
A: You could imagine that if the spring was not present, the block would slide down. This indicates that the friction f, which acts upward along the slope, is smaller than "g.sin(30) N".

2014-03-29 10:41:46 補充：
As such, the spring balance provides the necessary force to prevent the block sliding down. That is, spring force T = g.sin(30) - f

2014-03-29 10:42:37 補充：
The spring balance reading tells you the tension in the spring.

To begin with a diagram,
Normal reaction法向反作用力, Weight and (T + f) 作用於方塊移動的相反方向
Then, you may obtain an equation at equlibrum 9.81sin30 = T + f
In fact, 彈簧秤 measure 的只是spring 的tension張力
Therefore, f = 4.905 - 3.5 = 1.405 N
因為如果冇左條spring,個block會滑下