A solid bar of length L = 1.32 m has a mass m1 = 0.751 kg. The bar is fastened by a pivot at one end to a wall which is at an angle θ = 35.0 ° with respect to the horizontal. (in other words, the angle between the wall and the beam is 35 degrees).The bar is held horizontal by a vertical cord that is fastened to the bar at a distance xcord = 0.725 m from the wall. A mass m2 = 0.722 kg is suspended from the free end of the bar.

Question

A solid bar of length L = 1.32 m has a mass m1 = 0.751 kg. The bar is fastened by a pivot at one end to a wall which is at an angle θ = 35.0 ° with respect to the horizontal. (in other words, the angle between the wall and the beam is 35 degrees).The bar is held horizontal by a vertical cord that is fastened to the bar at a distance xcord = 0.725 m from the wall. A mass m2 = 0.722 kg is suspended from the free end of the bar.

What is the vertical component of the force exerted by the wall onto the beam? Assume the positive direction is upwards.
I know that the tension in the string is 20.2, and that the horizontal component is 0, but I'm having trouble creating the correction equation for the vertical component. My guess was Fy = m1g + mg2 - Tsin(55 degrees), but it turns out to be wrong. Help?

Answer 1

you "know" tension is 20.2 N?

Sum moments about the hinge

(m1*g*L/2+m2*g*L-T*L) sinTheta=0
L*g(m1/2+m2)=TL
T= g(.751/2+.722)
=9.81(1.09)=10.7

did I miss something?

Answer 2

For tension in the cable, I did 7.51N (1.32/2) + 7.22 (1.32-0.775)= 9.25. Then I took that number and divided it by 1.32, which equals 7.01. I then divided 7.01 by sin(55 degrees), which gives me a Tension of 20.2.

Answer 3

The force of gravity is also 10 N/kg for this problem.

Answer 4

To find the vertical component of the force exerted by the wall onto the beam, you need to consider the forces acting on the beam in the vertical direction and create an equation using Newton's second law.

First, let's analyze the forces acting on the beam in the vertical direction:

1. Weight of the beam (m1g): The mass of the beam (m1) multiplied by the acceleration due to gravity (g).
2. Weight of the suspended mass (mg2): The mass of the suspended mass (m2) multiplied by the acceleration due to gravity (g).
3. Tension in the string (T): The tension in the string is vertical and acts upwards to counterbalance the weight of the beam and the suspended mass.

Now, let's construct the equation:

Sum of vertical forces = m1g + mg2 - T * sin(θ)

Here's the step-by-step explanation for each term in the equation:

1. m1g: Represents the weight of the beam acting downwards.
2. mg2: Represents the weight of the suspended mass acting downwards.
3. T * sin(θ): The tension in the string has a vertical component that acts upwards. It can be found by multiplying the tension (T) by the sin of the angle (θ) between the string and the horizontal.

In this case, the angle θ is not given, so we need to find it using the given information:

θ = 90° - 35° (since the angle between the wall and the beam is 35°)
θ = 55°

Substituting the values back into the equation:

Sum of vertical forces = m1g + mg2 - T * sin(55°)

Now, plug in the given values for m1, m2, g, and T:

Sum of vertical forces = (0.751 kg)(9.8 m/s^2) + (0.722 kg)(9.8 m/s^2) - (20.2 N) * sin(55°)

Calculate the right side of the equation to find the vertical component of the force exerted by the wall onto the beam.