Torque Homework Help - Solve the Problem Now!

In summary, the minimum force required to tip a block of mass 28 kg, height 1.2 m, and length 0.69 m is 164.64 N.
  • #1
soccerj17
15
0
[SOLVED] Torque Homework Help

Homework Statement


Here's the problem and I need to submit the answer online tonight!

Consider the rectangular block of mass m = 28kg, height h = 1.2 m, length l = 0.69 m. A force F is applied horizontally at the upper edge. The acceleration of gravity is 9.8 m/s2. What is the minimum force required to start to tip the block?

Homework Equations


torque = force x distance
net torque = I*angular acceleration
So the force diagram would be mg pointing down from the center of mass, which is just the center of the box, normal pointing up from the bottom of the box, and force F pointing right from the top edge.

The Attempt at a Solution



Normal force wouldn't provide a torque because right? So it has to do with gravitational torque right? Well, gravitational torque would be mg * 0.6 m (half the height) which equals 164.64 Nm. What do I do from there to find the force needed to tip the box??
 
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  • #2
soccerj17 said:

Homework Statement


Here's the problem and I need to submit the answer online tonight!

Consider the rectangular block of mass m = 28kg, height h = 1.2 m, length l = 0.69 m. A force F is applied horizontally at the upper edge. The acceleration of gravity is 9.8 m/s2. What is the minimum force required to start to tip the block?


Homework Equations


So the force diagram would be mg pointing down from the center of mass, which is just the center of the box, normal pointing up from the bottom of the box, and force F pointing right from the top edge.


The Attempt at a Solution



Normal force wouldn't provide a torque because right? So it has to do with gravitational torque right? Well, gravitational torque would be mg * 0.6 m (half the height) which equals 164.64 Nm. What do I do from there to find the force needed to tip the box??
No, gravity acts dowm, you want to compute its torque about the bot right corner of the box, which is the pivot point at the point of tipping where all the normal force acts. You are then correct that the normal force provides no torque at this 'tipping' point. But you have incorrectly calculated the gravity torque about that corner. Once you correct it, now you can solve for F by equating the 2 torques.
 
  • #3
So the gravity torque isn't 164.64? I thought torque is force times perpendicular distance to the pivot point. Also I still don't see how to solve the problem. What 2 torques do I equate?
 
  • #4
soccerj17 said:
So the gravity torque isn't 164.64? I thought torque is force times perpendicular distance to the pivot point. Also I still don't see how to solve the problem. What 2 torques do I equate?
Yes, you may have your measurements mixd up, the block stands on its short end and is 1.2m tall; the weight thru the cg of the box is at 0.69/2 m perpendicular to the corner, right? Now your Force F is applied at the top end of the box, so what is its perpendicular distance from its line of action to the bot right corner?
 
  • #5
PhanthomJay said:
Yes, you may have your measurements mixd up, the block stands on its short end and is 1.2m tall; the weight thru the cg of the box is at 0.69/2 m perpendicular to the corner, right? Now your Force F is applied at the top end of the box, so what is its perpendicular distance from its line of action to the bot right corner?

So then torque gravity would be (28kg)*(9.81m/s2)*(.69/2m)? And Force F is .69 m from the bot right corner right?
 
  • #6
soccerj17 said:
So then torque gravity would be (28kg)*(9.81m/s2)*(.69/2m)? And Force F is .69 m from the bot right corner right?
you've got he gravity torque OK, bu the problem says the horizontal force is applied at the upper edge, that is, at the top of the box. So what's the perpendicular distance to the bottom? Have you drawn a sketch?
 
  • #7
Actually Force F is 1.2 m from the corner, right?
 
  • #8
Ok I got the answer right now. Thanks!
 

1. What is torque and why is it important?

Torque is the measure of a force's ability to cause rotation around a specific axis. It is important because it helps us understand how forces act on objects and how they can produce rotational motion.

2. How do I calculate torque?

Torque is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied. The formula is T = F x d, where T is torque, F is force, and d is distance.

3. What are the units of torque?

The units of torque are typically expressed in newton-meters (N·m) in the metric system and foot-pounds (ft·lb) in the imperial system. Other units, such as kilogram-force meters (kgf·m) and pound-force feet (lbf·ft), are also used in some cases.

4. Can you give an example of a torque problem?

Sure! Let's say you are trying to loosen a bolt with a wrench. The force you apply to the wrench is the magnitude of the torque, and the distance from the center of the bolt to the end of the wrench is the lever arm. The greater the force and the longer the lever arm, the greater the torque and the easier it will be to loosen the bolt.

5. How can I use torque to solve real-world problems?

Torque is used in a variety of real-world applications, such as in the design of machines and tools, the analysis of mechanical systems, and the development of new technologies. It can also be used to solve problems involving rotational motion, such as calculating the force needed to turn a crank or the power required to rotate a wheel.

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