# Torque Homework Help

[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??

Last edited:

PhanthomJay
Homework Helper
Gold Member

## 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.

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?

PhanthomJay
Homework Helper
Gold Member
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?

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?

PhanthomJay