Answer:
M = (2Tm)/g
Step-by-step explanation:
To determine the maximum mass that can be safely suspended in this manner, we need to analyze the forces acting on the box and apply the concept of equilibrium.
In this scenario, the tension in the rope at each end is acting upwards, counteracting the gravitational force pulling the box downwards.
Here are the steps to find the maximum mass, M:
1. Draw a free-body diagram for the box, representing all the forces acting on it. These forces include the gravitational force (mg) acting downwards and the tension (T) acting upwards at both ends of the rope.
2. Since the box is in equilibrium, the sum of all the forces in the vertical direction must be zero. Therefore, the tension at each end of the rope must be equal to half the weight of the box. This can be expressed as T = (1/2)mg.
3. We know that the maximum tension the rope can withstand without breaking is Tm. So, we can set T equal to Tm and solve for the maximum mass, M.
Tm = (1/2)mg
Simplifying the equation, we find:
M = (2Tm)/g
Therefore, the maximum mass that can be safely suspended in this manner is (2Tm)/g, where g is the acceleration due to gravity.
It is important to note that the above calculation assumes ideal conditions and neglects factors such as the mass of the rope and any other external forces. Additionally, the angle θ is not directly used in this calculation, but it affects the tension in the rope indirectly through trigonometry.