Question 1.

A girl whose mass is 55 kg stands on a spring weighing machine inside a lift. When the lift starts to ascend, its acceleration is 2 ms^-2. What will be the reading on the machine? (take g = 10 ms^-2) 

Solution:

Note: Two forces are acting in opposite direction on the girl. These are the force due to her weight (acting downwards), which equals

 mg = 55 x 10 = 550 Newtons

and the upwards force, the reaction, R, of the surface of the weighing machine on her. When the lift starts to move upward, the net force (R-mg) equals ma.

 From the formula for force, R - mg = ma

R = ma + mg

R = m (a + g)

     = 55 (2 + 10)

       = 660 Newtons

The force of Reaction, which is read by the machine is equal to 660 Newtons. 

In terms of mass, the reading on the machine will be

 ⁶⁶⁰/₁₀  = 66 kg

Question 2.

A spring balance, which is suspended from the roof of a lift, carries a mass of 1 kg at its free end. If the lift accelerates upwards at 2.5ms^-2, determine the reading on the spring balance if acceleration due to gravity (g) is 10m/s².

Solution:

Note: There are two different forces acting on the mass of 1 kg in opposite direction - the force due to its weight as a result of the force of gravity, mg, and the upward force on it from the spring balance (F).

The net value of these two forces causes the lift to accelerate upward.

Therefore, using the formula of force,

F - mg = ma

F = ma + mg

F = m(a + g)

    = (2.5 + 10)

F = 12.5N

The reading on the spring balance is equal to 12.5 Newtons

Question 3.

An elevator of mass 4800 kg is supported by a cable which can safely withstand a maximum tension of 60 000 N. The maximum upward acceleration the elevator can have, taking acceleration due to gravity as 10 ms^-2, is?

Solution:

Two forces are acting on the elevator: 1. the weight of the elevator, given by the force formula,

                               W = mg

                              W = 4800 x 10

                                  = 48000 Newtons

1. The tension (T) in the cable acting upwards and given as 60 000N when the elevator starts to move upward, the net force equals the difference in the two forces.

Using the formula of force,

T - mg = ma 

60 000 - 48 000  = 4800a

12 000 = 48 00a

a = 12 000/4800 = 2.5

The maximum upward acceleration the elevator can have is 2.5 m/s²

Question 4.

A body of mass 2 kg is suspended from the ceiling of a lift with a light insensible string. If the lift moves upwards with acceleration of 2 ms^-2, calculate the magnitude of the tension in the string. (g = 10 ms^-2).

Solution:

There are two forces acting on the body: vertical upward force, which is same as the tension (T) in the string, and downward force, which is due to its weight (mg).

Since the lift is moving upwards, the net force acting on the body is given by the equation:

T - mg = ma

Given: mass of the body m = 2 kg, acceleration a = 2 ms^-2, acceleration due to gravity g = 10 ms-², tension in the string T = ?

Therefore,

T = ma + mg

T = m (a + g)

   = 2 (2 + 10)

= 2 x 12

   = 24 N

The tension in the string is 24 Newtons.

Question 5.

A 1000 kg elevator is descending vertically with an acceleration of 1.0 ms^-2. If the acceleration due to gravity is 10.0 ms^-2, the tension in the suspending cable is? 

Solution:

Two forces act on the elevator: the vertical upward force (it actually acts downwards in this case), which is also the tension (T), and the weight of the elevator, mg.

Since the elevator is moving downwards, the same direction as its weight due to the force of gravity acting on it, the net force acting on it is

T + mg = ma

T + 1000 x 10 = 1000 x 1

T + 10000 = 1000

T = - 9000 N (the negative sign shows that the tension in the cable is acting downwards) 

Therefore, the tension in the cable is 9000 Newtons.

See more calculations of force based on apparent weight.