Help: A 2 meter long, uniform bar with a mass of 46 kg is suspended from a pin so it can rotate freely. It is supported by a cable half a meter from the end. Find the tension force in the cable and the magnitude and direction of the reaction force at the wall for the bar to move
fw+T=46*9.8 vertical forces.
summing now about the wall..
46(9.8)*1=T*2
tension=23*9.8
fw= First equation=you do it.
To find the tension force in the cable and the reaction force at the wall for the bar to move, you can use static equilibrium equations. In static equilibrium, the sum of all forces acting on an object is zero, and the sum of all torques (or moments) acting on the object is also zero.
Firstly, let's consider the forces acting on the bar. There are two forces: the tension force in the cable and the reaction force at the wall. The tension force in the cable pulls upward, while the reaction force at the wall pushes inward (perpendicular to the wall).
Now, let's analyze the torques acting on the bar. Since the bar is in rotational equilibrium, the sum of all torques acting on it must be zero. The torques are caused by the gravitational force acting downwards and the tension force and reaction force acting at different positions on the bar.
To solve the problem, we need to set up the equations for the sum of the forces and the sum of the torques acting on the bar.
Let's denote:
- T as the tension force in the cable
- R as the magnitude of the reaction force at the wall
- L as the length of the bar (2 meters)
- d as the distance from the pivot point to the cable (0.5 meter)
Now, let's write down the equilibrium equations:
Sum of forces in the vertical direction (y-direction):
T - mg = 0
Sum of torques about the pivot point:
(R * L) - (T * d) - (m * g * L/2) = 0
In the first equation, mg represents the gravitational force acting downwards, which is equal to the weight of the bar (mass * gravitational acceleration ≈ 46 kg * 9.8 m/s^2).
Simplifying the second equation, we can rewrite it as:
R * L - T * d - mg * L/2 = 0
Now, we can solve these equations simultaneously to find the values of T and R.
Substituting T = mg into the second equation, we have:
R * L - mg * d - mg * L/2 = 0
R * L - mg * L/2 = mg * d
Simplifying further:
R = (mg * L/2 - mg * d) / L
Now, substitute the known values:
R = (46 kg * 9.8 m/s^2 * 2 m/2 - 46 kg * 9.8 m/s^2 * 0.5 m) / 2 m
R = (46 kg * 9.8 m/s^2 - 46 kg * 9.8 m/s^2 * 0.5) / 2
R = (46 kg * 9.8 m/s^2 * (1 - 0.5)) / 2
R = (46 kg * 9.8 m/s^2 * 0.5) / 2
Now, calculate R:
R = 225.68 N
So, the magnitude of the reaction force at the wall is approximately 225.68 Newtons.
To find the tension force in the cable, substitute T = mg into the first equation:
T = mg
T = 46 kg * 9.8 m/s^2
T = 450.8 N
Therefore, the tension force in the cable is approximately 450.8 Newtons.
Note: The positive direction of the reaction force will be towards the wall, and the positive direction of the tension force will be upward from the cable.