# Turning effects of force 5

## Homework Statement

A forklift truck of weight W has dimensions shown in the figure below. Its centre of gravity G is mid-way between the front and the rear axles.

When a uniform block of
weight w (w < W) is placed on the front forks as shown, how the load on the rear axle will change?

## Homework Equations

Will the truck's center of gravity G change?

## The Attempt at a Solution

I tried to use M = F.d, but I don't know which items are constant, and which ones change.

haruspex
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When taking moments, it helps to choose the axis you take moments about carefully. Where would you think is the best place here?

longnh
BvU
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M = F d is the relevant equation indeed. Do pay attention to the signs !

General remark: before doing 'real work', you can skip answers (c) (because that's not a load but a number) and (d) (it certainly won't increase).

In fact, you can also distinguish between (a) and (b) in a lazy way: what happens with the load if w comes close to W ?

To help you with your 'relevant equation' : G stays in the same place for the truck itself. So there are two torques to consider: from the truck and from the load.

longnh
Thank you guys for tipping.

In a lazy way, the answer should be (a). w/2 but I cannot find a solution to that result.

Consider G stays unchanged.

At early state: M1=M2
<=> F1.d = F2.d
<=> W1 = W2 = W/2

When that block is loaded, we have M'1=M'2
<=> F1.d + F.2d = F'2.d
<=> (W/2).d + w.2d = (W'/2).d
<=> :L:L:L

haruspex
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2020 Award
Thank you guys for tipping.

In a lazy way, the answer should be (a). w/2 but I cannot find a solution to that result.

Consider G stays unchanged.

At early state: M1=M2
<=> F1.d = F2.d
<=> W1 = W2 = W/2

When that block is loaded, we have M'1=M'2
<=> F1.d + F.2d = F'2.d
<=> (W/2).d + w.2d = (W'/2).d
<=> :L:L:L
Please state what point you are taking moments about. If it is the centre of mass of the truck then you have a distance wrong.
As BvU posted, be careful with signs. Best way is to pick a direction (left or right) as positive and measure each distance from the axis to the force using the appropraite sign. Doing things that way you write that the sum of moments is zero.
Another approach is just to stop and think which moments are clockwise and which are anticlockwise and write that the sum of the clockwise equals the sum of the anticlockwise.

longnh
Yes, haruspex, because the truck didn't rotate, sum of counterclockwise moments (front wheels) equals to clockwise moment (at rear wheels). So I have M'1 = M'2 written above.

If I choose an axis with positive direction from left to right, then ∑M = 0 ⇔ (W'/2).d - w.2d - (W/2).d, which is the same as my equation above.

The equation ∑M = M1 - M2 and M1 = M2 are alternative.

The thing that matters is if that block is loaded, is the whole weight of the truck changed, too? Or we consider that w doesn't affect W?

Please state what point you are taking moments about. If it is the centre of mass of the truck then you have a distance wrong.
As BvU posted, be careful with signs. Best way is to pick a direction (left or right) as positive and measure each distance from the axis to the force using the appropraite sign. Doing things that way you write that the sum of moments is zero.
Another approach is just to stop and think which moments are clockwise and which are anticlockwise and write that the sum of the clockwise equals the sum of the anticlockwise.

ehild
Homework Helper
The thing that matters is if that block is loaded, is the whole weight of the truck changed, too? Or we consider that w doesn't affect W?

Weight means mass times g here, the force of gravity. The weight of the truck itself does not change if load is added.

Consider the moments (torques) of all forces acting on the truck+load system. They are the forces of gravity W and w acting at the centres of mass, and the upward normal forces acting between the ground and the wheels. Load on the rear axle is equal in magnitude with the normal force acting on the real wheel. Choose the front axle and write the moments of all forces with respect to it.

haruspex
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Yes, haruspex, because the truck didn't rotate, sum of counterclockwise moments (front wheels) equals to clockwise moment (at rear wheels). So I have M'1 = M'2 written above.

If I choose an axis with positive direction from left to right, then ∑M = 0 ⇔ (W'/2).d - w.2d - (W/2).d, which is the same as my equation above.

The equation ∑M = M1 - M2 and M1 = M2 are alternative.

The thing that matters is if that block is loaded, is the whole weight of the truck changed, too? Or we consider that w doesn't affect W?
You're still not saying what point you are taking moments about. Is it the mass centre of the (unloaded) truck? The front wheels?

haruspex
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Yes, haruspex, because the truck didn't rotate, sum of counterclockwise moments (front wheels) equals to clockwise moment (at rear wheels). So I have M'1 = M'2 written above.

If I choose an axis with positive direction from left to right, then ∑M = 0 ⇔ (W'/2).d - w.2d - (W/2).d, which is the same as my equation above.

The equation ∑M = M1 - M2 and M1 = M2 are alternative.

The thing that matters is if that block is loaded, is the whole weight of the truck changed, too? Or we consider that w doesn't affect W?
You're still not saying what point you are taking moments about. Is it the mass centre of the (unloaded) truck? The front wheels?

You're still not saying what point you are taking moments about. Is it the mass centre of the (unloaded) truck? The front wheels?

haruspex
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2020 Award
Thankyou.
There are a couple of problems with the equation you get in post #4.
First, there are four forces acting on the truck: gravity of the truck, the weight of the load w, and the normal force at each wheel.
Gravity on the truck has no momentabout the truck's mass centre, so you can ignore that. The load on the rear axle is equal to the normal force from the road there. Therefore your F1, F2 should be the normal forces from the road. In that light, reconsider the signs in the equation.
Next, you must not assume that one of those normal forces is unchanged by the load.
By the way, it might be a little quicker if you take moments about the front axle instead. That will avoid bringing the normal force on the front axle into the equation.

longnh
I might see it now.

I used to think that the pivot was that G point, but I was wrong. If the truck was rotated, the pivot would be the front axle.

At the front axle point: w.d = F2.2d ⇔ F2 = w/2

Thank you!

Thankyou.
There are a couple of problems with the equation you get in post #4.
First, there are four forces acting on the truck: gravity of the truck, the weight of the load w, and the normal force at each wheel.
Gravity on the truck has no momentabout the truck's mass centre, so you can ignore that. The load on the rear axle is equal to the normal force from the road there. Therefore your F1, F2 should be the normal forces from the road. In that light, reconsider the signs in the equation.
Next, you must not assume that one of those normal forces is unchanged by the load.
By the way, it might be a little quicker if you take moments about the front axle instead. That will avoid bringing the normal force on the front axle into the equation.

haruspex
Homework Helper
Gold Member
2020 Award
I might see it now.

I used to think that the pivot was that G point, but I was wrong. If the truck was rotated, the pivot would be the front axle.
You can make the pivot point for the equation wheever you like, but some choices are more convenient than others.
Using the mass centre works, but the equation involves w and the normal force at each axle. You then need to use the statics equation for the sum of vertical forces to get a second equation involving the two normal forces at the axles and manipulate the equations to eliminate the one of no interest (the normal force at front axle).
At the front axle point: w.d = F2.2d ⇔ F2 = w/2
That depends what you mean by F2.
If we write F for the new normal force at the rear axle and take moments about the front axle we get an equation involving w, F and W.

longnh