In Fig. 5-27, a block attached to a chord is resting on an inclined plane. Let the mass of the block be 8.5 kg and the angle `theta` be `30^(@)`.
(a) Find the tension in the cord and normal force acting on the block.
(b) If the cord is cut, find the magnitude of the resulting acceleration of the block.

Figure 5-27 A block attached to a chord resting on an incline.

Explanation

(1) We apply Newton.s second law to solve for the tension in the cord and the normal force on the block.
(2) The free-body diagram of the problem is shown in the Fig. 5-28. Since the acceleration of the block is zero, the components of Newton.s second law equation yield
`T-"mg "sin theta=0" "(5-39)`
`F_(N)="mg "cos theta=0," "(5-40)`
where T is the tension in the cord, and `F_(N)` is the normal force on the block.

Figure 5.28 Free body diagram of the given problem.
Calculation: Solving Eq. (5-39) for the tension in the string, we find
`T="mg "sin theta=(8.5" kg")(9.8" m"//"s"^(2))sin30^(@)`
`=42" N".`
We solve Eq. (5-40) above for the normal force `F_(N)`.
`F_(N)="mg "cos theta=(8.5" kg")(9.8" m"//"s"^(2))cos30^(@)`
`=72" N"`.
(b) If the cord is cut, find the magnitude of the resulting acceleration of the block.
Calculation: When the cord is cut, it no longer exerts a force on the block and the block accelerates. The x component of the second law becomes -mg `sin theta=ma`, so the acceleration becomes
`a=-g sin theta=-(9.8" m"//"s"^(2))sin 30^(@)`
`=-4.9" m"//"s"^(2)`.
The negative sign indicates the acceleration is down the plane. The magnitude of the acceleration is `4.9" m"//"s"^(2)`.
Note: The normal force `F_(N)` on the block must be equal to mg `cos theta` so that the block is in contact with the surface of the incline at all time. When the cord is cut, the block has an acceleration `a=-g sin theta`, which in the limit `theta to 90^(@)` becomes `-g`, as in the case of a free fall.