A weight (with a mass of 71 kg) is suspended from a point near the right-hand end of a uniform boom with a mass of 51 kg . To support the uniform boom a cable runs from this same
point to a wall (the left-hand vertical coordinate y=7 x=6) and by a pivot on the same wall at an elevation of 3 m.
Calculate the tension T in the cable. The
acceleration of gravity is 9.8 m/s2 .
Answer in units of N
To calculate the tension in the cable, we need to consider the forces acting on the boom.
1. Weight of the boom (Fboom): Fboom = mass of the boom * acceleration due to gravity
Given that the mass of the boom is 51 kg, and the acceleration due to gravity is 9.8 m/s^2,
Fboom = 51 kg * 9.8 m/s^2
2. Weight of the weight (Fweight): Fweight = mass of the weight * acceleration due to gravity
Given that the mass of the weight is 71 kg, and the acceleration due to gravity is 9.8 m/s^2,
Fweight = 71 kg * 9.8 m/s^2
3. Pivot force (Fpivot): Fpivot acts in the opposite direction of Fboom and Fweight to keep the boom in equilibrium.
Fpivot = Fboom + Fweight
4. Tension in the cable (T): T applies an upward force to balance Fpivot and the horizontal forces.
T = Fpivot
Now, let's calculate the values step by step.
Fboom = 51 kg * 9.8 m/s^2 = 499.8 N
Fweight = 71 kg * 9.8 m/s^2 = 696.8 N
Fpivot = Fboom + Fweight = 499.8 N + 696.8 N = 1196.6 N
Therefore, the tension in the cable (T) is 1196.6 N.