Question

A bedroom bureau with a mass of 36 kg, including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is 0.60, what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with 18 kg mass, are removed before the bureau is pushed, what is the new minimum magnitude?

Answer 1

a) 211.7 N b) 105.8 N

Step-by-step explanation:

a)

• The friction force, adopts any value needed to meet the Newton's 2nd Law, so, while the bureau is at rest, the friction force is equal and opposite to the applied force.
• This friction force achieves its maximum value, when the applied force, is just on the verge of getting the object to move.
• At that moment, the friction force can be calculated as follows:

`F_(f) = \mu_(s) * F_(n) (1)`

• where μs = coefficient of static friction = 0.60, and Fₙ = Normal Force.
• The normal force, is always perpendicular to the surface on which the object is placed, and always aims upward.
• As the friction force, the normal force can adopt any value needed to prevent the object to accelerate in the vertical direction, going through the surface under the influence of gravity.
• In this case, if the floor is level, we have the following equation:

`F_(n) = m*g`

• Replacing in (1) , we have:

`F_(f) = \mu_(s) * m* g = 0.60* 36 kg * 9.8 m/s2 = 211. 7 N`

• As we have already told, this force must be equal in magnitude to the minimum applied force that it is needed to apply to start the bureau moving.
• So, Fmin = 211.7 N

b)

• If we reduce the value of m, the new minimum force needed will be as follows:

`F = \mu_(s) * m* g = 0.60* 18 kg * 9.8 m/s2 = 105.8 N`