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Question: Block A of weight \(100{\text{N}}\) reset on a frictionless inclined plane of slope angle \({30^\cir...

Block A of weight 100N100{\text{N}} reset on a frictionless inclined plane of slope angle 30{30^\circ } as shown in the figure. A flexible cord attached to AA passes over a frictionless pulley and is connected to a block BB of weight WW. Find the weight WW of which the system is in equilibrium.

Explanation

Solution

We have to know how all the forces are acting on this system at every point. We can then continue to break every force vector into its components, which will give us a better overview of this problem. After that, we will balance every force component and will try to find out our desired answer.

Complete step by step solution:
We will break every force vectors into its components and will add the tension on the string, and redraw our given figure as following-

In the above figure, we have-
TT is the tension in the string
WW is the weight of the block B{\text{B}}
NN is the normal force acting on the block A{\text{A}} by the surface
mgmg is the weight of the block A{\text{A}}
Given that the weight of the block AA is 100N100{\text{N}} i.e. mg=100Nmg = 100{\text{N}}
All the forces are acting in the same direction, as shown by arrows in the figure. We break mgmg into its components, and we have mgcos30mg\cos {30^\circ } to balance the vector NN and mgsin30mg\sin 30^\circ to balance TT. Also, the vector WW is balancing TT on the other side.
Hence in the equilibrium of the system, we have-
T=WT = W ………..(1)(1)
T=mgsin30T = mg\sin {30^\circ }……....(2)(2)
By comparing equations (1)(1) and (2)(2), we get-
W=mgsin30W = mg\sin {30^\circ }
We put mg=100Nmg = 100{\text{N}} and sin30=12\sin {30^\circ } = \dfrac{1}{2}-
W=100×12N\therefore W = 100 \times \dfrac{1}{2}{\text{N}}
W=50N\Rightarrow W = 50{\text{N}}

Therefore, the weight of the block BB is 50N50{\text{N}}.

Note: This problem is a special case of the most generic problem. The tension at the two parts of the string is identical because the pulley and the inclined surface are frictionless. Furthermore, the sting is taken to be as massless here. In the generic problem, one must take account of the other forces that are involved.