A block Q Of mass 70 kg is at rest on a table it is connected to block P By means two light inextensible strings knotted at S .A Third string is arranged in such a way that the string connecting block Q is horizontal.
Calculate the maximum weight of block P for which block Q will just begin to slip
To calculate the maximum weight of block P for which block Q will just begin to slip, we need to consider the forces acting on block Q. There are two forces acting on block Q: the weight (mg) acting vertically downwards and the tension in the string (T) acting horizontally towards the right. The tension in the string can be broken down into two components: one component acting vertically upwards (Tsinθ), which balances the weight of block Q, and the other component acting horizontally towards the right (Tcosθ), which can cause the block to slip.
To find the angle θ, we can look at the geometry of the system. The string connecting block Q is horizontal, which means the angle between the string and the vertical direction is 90 degrees. Therefore, the angle between the string and the horizontal direction (θ) is also 90 degrees.
Since the angle θ is 90 degrees, the horizontal component of tension (Tcosθ) becomes zero. Therefore, the tension in the string (T) will only balance the weight of block Q, which means the maximum weight of block P for which block Q will just begin to slip is equal to the weight of block Q.
Weight of block Q = mass of block Q * acceleration due to gravity
Weight of block Q = 70 kg * 9.8 m/s^2
Weight of block Q = 686 N
Therefore, the maximum weight of block P for which block Q will just begin to slip is 686 N.