Calculating Tension on 500kg Beam Supported by Cable

• yankees26an
In summary: Thanks :)In summary, the tension in the cord is equivalent to the sum of the torques from all the forces acting on the beam at the pivot.
yankees26an

Homework Statement

A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
the cable?

Torque = r * F

The Attempt at a Solution

Using pivot point at the end of the beam, the torque is 0.

weight acting down on the beam is mg*l/2*sin(35). L is not givin not sure where to go from there.

yankees26an said:

Homework Statement

A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
the cable?

Torque = r * F

The Attempt at a Solution

Using pivot point at the end of the beam, the sum of the torques from all forces is 0.

torque from weight acting down on the beam is mg*l/2*sin(35), clockwise. L is not givin not sure where to go from there.
What's the torque from the cord tension about the pivot?

PhanthomJay said:
What's the torque from the cord tension about the pivot?

you mean from the top of the beam to the bottom?

yankees26an said:
you mean from the top of the beam to the bottom?
yes.

PhanthomJay said:
yes.

Not sure how to find it.

yankees26an said:
Not sure how to find it.
You correctly identified the torque about the pivot from the weight force as mgL/2sintheta, that is Torque = r X F = rFsintheta, where F in this case is mg, and r is L/2, and theta is 35 degrees(the angle between the force and position vector). Now do the same for the torque about the pivot for the unknown horizontal tension force (call it T) at the top of the beam, that is, Torque from T = r X F = rFsintheta, where r is L, T is itself, and theta , the angle between T and L , is ?.. Solve for T by summing these 2 torques and setting that sum equal to 0. The 'L" term should cancel out.

mg*L/2*sin(35) = T*L*sin(55)

mg/2*sin(35) = T*sin(55)

(mg/2*sin(35))/sin(55) = T

T = 1.72 kN

Thanks :)

1. How do you calculate the tension on a 500kg beam supported by a cable?

The tension on a 500kg beam supported by a cable can be calculated using the formula T = mg + ma, where T is the tension, m is the mass of the beam, g is the acceleration due to gravity (9.8m/s^2), and a is the acceleration of the beam. The acceleration of the beam can be calculated using Newton's second law, F = ma, where F is the net force acting on the beam.

2. What factors affect the tension on a 500kg beam supported by a cable?

The factors that affect the tension on a 500kg beam supported by a cable include the mass of the beam, the angle of the cable, the length and strength of the cable, and the forces acting on the beam (such as gravity or external loads).

3. How does the angle of the cable affect the tension on a 500kg beam?

The angle of the cable affects the tension on a 500kg beam by changing the direction of the force applied on the beam. As the angle of the cable increases, the horizontal component of the tension decreases, while the vertical component increases. This can result in a higher tension on the cable and potentially cause it to break.

4. Why is it important to calculate the tension on a 500kg beam supported by a cable?

It is important to calculate the tension on a 500kg beam supported by a cable because it ensures the structural integrity and safety of the beam. If the tension exceeds the weight-bearing capacity of the cable, it could result in the cable breaking and causing damage or injury.

5. Can tension on a 500kg beam supported by a cable be negative?

No, tension cannot be negative. Tension is a type of force and is always positive or zero. If the calculation results in a negative value, it means that the cable is under compression rather than tension. In this case, the cable may be at risk of buckling or breaking, and alternative support methods should be considered.

• Introductory Physics Homework Help
Replies
13
Views
369
• Introductory Physics Homework Help
Replies
10
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
18
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K