Question

The weight of the crate causes a clockwise moment of 13 800 Nm about the centre of the front wheel of the fork-lift truck. (b) The weight of the fork-lift truck and driver cause an anticlockwise moment. What is the minimum size of the anticlockwise moment needed so that the fork-lift truck does not topple over?

Answer 1

Answer:

anser is a

Explanation:

Answer 2

a) 29,900 J

b) 13,800 Nm

Explanation:

a)

The work done by a force on an object is given by:

$W=Fd$

where

F is the magnitude of the force

d is the displacement of the object

In this problem:

The force applied on the crate must be equal (at least) to the weight of the crate. Therefore, in this case,

F = W = 11,500 N

The distance through which the crate has been lifted is

d = 2.6 m

Therefore, the work done to lift the crate is:

$W=(11500)(2.6)=29,900 J$

b)

Here we are said that the weight of the fork-lift truck and driver cause an anticlockwise moment.

When there is an unbalanced moment, the object starts to rotate around a certain pivotal point.

In this case, we want to find the minimum size of the anticlockwise moment needed in order for the fork-lift truck not to rotate.

In order for this to happen, the net overall moment on the truck must be zero: this means that the anticlockwise moment must be equal to the clockwise moment,

$M_A=M_C$

In this case, the clockwise moment is

$M_C=13,800 Nm$

Therefore, the anticlockwise moment must be equal:

$M_A=13,800 Nm$