3.7k views
5 votes
A contractor is pushing a stove across a kitchen floor with a constant velocity of 18 cm/s [fwd]. The contractor is exerting a constant horizontal force of 85 N [fwd]. The force of gravity on the stove is 447 N [down].

Determine the normal force (FN ) and the force of friction (Ff ) acting on the stove.
Determine the total force applied by the floor (Ffloor) on the stove.

User Jalanga
by
7.4k points

1 Answer

3 votes

Answer:

Step-by-step explanation:

Since the stove is moving with a constant velocity, we know that the net force on the stove is zero. Therefore, the force of friction acting on the stove must be equal in magnitude and opposite in direction to the horizontal force being applied by the contractor. We can use Newton's second law to solve for the normal force and force of friction:

ΣF = ma

where ΣF is the net force, m is the mass of the stove, and a is the acceleration of the stove (which is zero in this case).

First, we need to convert the velocity to m/s and the forces to Newtons (N):

18 cm/s = 0.18 m/s

85 N [fwd] - force applied by contractor

447 N [down] - force of gravity on the stove

Now we can solve for the normal force:

ΣFy = 0 (since the stove is not accelerating in the y-direction)

FN - 447 N = 0

FN = 447 N

Therefore, the normal force acting on the stove is 447 N.

Next, we can solve for the force of friction:

ΣFx = 0 (since the stove is moving at a constant velocity)

Ff - 85 N = 0

Ff = 85 N [bkwd]

Therefore, the force of friction acting on the stove is 85 N [bkwd].

Finally, we can solve for the total force applied by the floor:

ΣF = ma = 0 (since the stove is not accelerating)

Ffloor - 85 N - 447 N = 0

Ffloor = 532 N [up]

Therefore, the total force applied by the floor on the stove is 532 N [up].

User Enzotib
by
8.6k points