# Questions regarding a workman's toolbox

A workman carries a sheet of metal up to the roof of a skyscraper by transporting it in the building's elevator. Inside the elevator, the metal sits at an angle of 26o from the horizontal. The workman absentmindedly places his toolbox on the smooth metal surface, and it begins to slide with negligible friction.

a. Draw a free-body diagram for the toolbox.

b. Before the elevator starts to move, what is the acceleration of the toolbox along the

metal surface?

c. When the elevator is accelerating upward at 0.40 g (that is, at 40% of the acceleration due to gravity), the toolbox feels a normal force of 130 N. What is the mass of the toolbox?

d. If the toolbox feels a normal force of 80 N, find the acceleration of the elevator.

e. Could the elevator accelerate in such a way that the toolbox will not accelerate along the metal surface at all? If so, find the acceleration of the elevator; if not, explain why not.

https://brainmass.com/physics/acceleration/questions-regarding-workmans-toolbox-398389

## SOLUTION This solution is **FREE** courtesy of BrainMass!

A workman carries a sheet of metal up to the roof of a skyscraper by transporting it in the building's elevator. Inside the elevator, the metal sits at an angle of 26O from the horizontal. The workman absentmindedly places his toolbox on the smooth metal surface, and it begins to slide with negligible friction.

a. Draw a free-body diagram for the toolbox.

b. Before the elevator starts to move, what is the acceleration of the toolbox along the

metal surface?

c. When the elevator is accelerating upward at 0.40 g (that is, at 40% of the acceleration due to gravity), the toolbox feels a normal force of 130 N. What is the mass of the toolbox?

d. If the toolbox feels a normal force of 80 N, find the acceleration of the elevator.

e. Could the elevator accelerate in such a way that the toolbox will not accelerate along the metal surface at all? If so, find the acceleration of the elevator; if not, explain why not.

Solution:

a) Draw a free-body diagram for the toolbox.

See attached file for equations.

Forces acting on the tool box are: i) its weight mg which has been resolved into components mgsinÎ¸ and mgcosÎ¸ and ii) normal reaction of the metal sheet.

b) Before the elevator starts to move, what is the acceleration of the toolbox along the

metal surface?

The force component acting on the tool box along the metal surface = mgsinÎ¸

Acceleration of the tool box = Force/Mass = mgsinÎ¸/m = gsinÎ¸ = 9.8sin26O = 4.3 m/s2

c) When the elevator is accelerating upward at 0.40 g (that is, at 40% of the acceleration due to gravity), the toolbox feels a normal force of 130 N. What is the mass of the toolbox?

As the elevator starts to accelerate upward, the forces acting on it undergo changes. We draw the modified force diagram hereunder:

See attached file for calculations.

d) If the toolbox feels a normal force of 80 N, find the acceleration of the elevator.

Substituting N = 80 N, m = 8.52 kg, Î¸ = 26O, g = 9.8 m/s2 in (1) we get:

a = (80cos26O - 8.52x9.8)/8.52 = - 1.36 m/s2 (the elevator is accelerating downward)

e) Could the elevator accelerate in such a way that the toolbox will not accelerate along the metal surface at all? If so, find the acceleration of the elevator; if not, explain why not.

As the elevator accelerates up or down, magnitude of normal reaction N varies to satisfy equation (1). However, N being perpendicular to the metal sheet, it has no component along the metal sheet. Hence, variation of N has no effect on the net force acting on the box along the metal sheet. Net force along the metal sheet is constant at mgsinÎ¸. Hence, box's acceleration down the metal sheet is unaffected by the acceleration of the elevator.

(Note: In the above treatment we have assumed that friction is zero. However, in case of a rough surface, there will be a frictional force acting upward along the surface; the frictional force being directly proportional to the normal reaction N (f = Î¼N). Hence, net force on the box along the surface is given by: mgsinÎ¸ - Î¼N. In this case there will be a value of acceleration of the elevator at which N will satisfy the equation mgsinÎ¸ = Î¼N resulting is net zero force on the box and consequently zero acceleration down the plane.)

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