# Mechanics Question

A lift of mass Mkg is being raised by a vertical cable attached to the top of the lift. A person of mass mkg stands on the floor inside the lift, as shown in Figure 1. The lift ascends vertically with constant acceleration 1.4 m s. The tension in the cable is 2800 N and the person experiences a constant normal reaction of magnitude 560 N from the floor of the lift. The cable is modelled as being light and inextensible, the person is modelled as a particle and air resistance is negligible.By writing an equation of motion for the person only, and an equation of motion for the lift only, find the value of M and the value of m.
So you need to apply Newton's second law, resultant force =mass x acceleration, on the person and lift separately.
You should get 2 equations to solve simultaneously.
Original post by GhostWalker123
A lift of mass Mkg is being raised by a vertical cable attached to the top of the lift. A person of mass mkg stands on the floor inside the lift, as shown in Figure 1. The lift ascends vertically with constant acceleration 1.4 m s. The tension in the cable is 2800 N and the person experiences a constant normal reaction of magnitude 560 N from the floor of the lift. The cable is modelled as being light and inextensible, the person is modelled as a particle and air resistance is negligible.By writing an equation of motion for the person only, and an equation of motion for the lift only, find the value of M and the value of m.

It would first be helpful to draw a diagram when attempting any mechanical problem such as this.

Motion of lift:
F=Ma
2800-Mg=M(1.4)

g=-9,8
Rearrange for M: M=250kg

Motion of person:
F=ma
560-mg=m(1.4)

g=-9,8
Rearrange for m:

Rearrange for m: m=50kg
Original post by Drogonmeister
It would first be helpful to draw a diagram when attempting any mechanical problem such as this.

Motion of lift:
F=Ma
2800-Mg=M(1.4)

g=-9,8
Rearrange for M: M=250kg

Motion of person:
F=ma
560-mg=m(1.4)

g=-9,8
Rearrange for m:

Rearrange for m: m=50kg

This is wrong for the motion of the lift u have to consider newtons 3rd law and hence there should be a downwards reaction force of 560 acting on the lift from the man. Even tho we ignore the man he still exerts a force on the floor of the lift and this force will be the same as that ofbthe lift exerting 560 to the man. The reason they dont cancel is because 1 force only acts of lift and the other force only acts on man. So ur calculations shld be 2800-560-Mg=M(1.4) and so on
Original post by Drogonmeister
It would first be helpful to draw a diagram when attempting any mechanical problem such as this.