A box having a mass of 50kg is dragged across a horizontal floor by means of a rope tied on the front of it. The coefficient of friction between the box and the floor is .3. If the angle between the rope and the floor is 30 degrees, what force must be excerted on the rope to move the box at uniform speed?

To determine the force required to move the box at uniform speed, we need to consider the gravitational force and the frictional force acting on the box.

Step 1: Calculate the gravitational force:
The gravitational force can be calculated using the formula: F_gravity = mass × gravitational acceleration
Given that the mass of the box is 50 kg and the gravitational acceleration is approximately 9.8 m/s², we can calculate the gravitational force:
F_gravity = 50 kg × 9.8 m/s²
F_gravity = 490 Newtons

Step 2: Calculate the frictional force:
The frictional force can be calculated using the formula: F_friction = coefficient of friction × normal force
The angle between the rope and the floor is 30 degrees. Therefore, the normal force is equal to the gravitational force times the cosine of the angle:
Normal force = F_gravity × cos(angle)
Normal force = 490 N × cos(30 degrees)
Normal force = 490 N × 0.866
Normal force = 424.94 Newtons

Frictional force = coefficient of friction × normal force
Frictional force = 0.3 × 424.94 N
Frictional force = 127.48 Newtons

Step 3: Calculate the force applied to the rope:
Since the box is moving at a uniform speed, the force applied to the rope must be equal to the frictional force in the opposite direction.
Therefore, the force applied to the rope is 127.48 Newtons.

So, the force that must be exerted on the rope to move the box at uniform speed is 127.48 Newtons.

To find the force that must be exerted on the rope to move the box at a uniform speed, we need to consider the forces acting on the box.

First, we have the gravitational force pulling the box downward, which can be calculated using the formula:

F_gravity = mass * acceleration due to gravity
F_gravity = 50 kg * 9.8 m/s^2 (acceleration due to gravity)
F_gravity = 490 N

Next, we have the normal force exerted by the floor on the box. Since the box is on a horizontal surface and there is no vertical acceleration, the normal force is equal in magnitude but opposite in direction to the gravitational force, i.e., 490 N.

The force of friction can be calculated using the formula:

F_friction = coefficient of friction * normal force
F_friction = 0.3 * 490 N (coefficient of friction * normal force)
F_friction = 147 N

The force exerted on the rope can be resolved into two components: one parallel to the floor (horizontal component) and one perpendicular to the floor (vertical component).

The horizontal component of the force exerted on the rope is responsible for overcoming the force of friction:

F_horizontal = F_friction = 147 N

The vertical component of the force exerted on the rope can be calculated using trigonometry, as the angle between the rope and the floor is given as 30 degrees:

F_vertical = Force on the rope * sin(angle)
F_vertical = Force on the rope * sin(30 degrees)

Since the box is being dragged at a uniform speed, we know that the net force acting on the box is zero. This means that the horizontal component of the force should be equal to the force of friction:

F_horizontal = F_friction = 147 N

Now we can calculate the force exerted on the rope:

F_horizontal = Force on the rope * cos(angle)
147 N = Force on the rope * cos(30 degrees)

Rearranging the equation to solve for the force on the rope:

Force on the rope = 147 N / cos(30 degrees)

Using a calculator, you can find that cos(30 degrees) is approximately 0.866:

Force on the rope = 147 N / 0.866
Force on the rope ≈ 169.42 N

Therefore, a force of approximately 169.42 Newtons must be exerted on the rope to move the box at a uniform speed.

75

Let the froce be F.

The horizontal component is Fx = F cos 30

That must equal the weight times 0.3.

F cos30 = 50 g * (0.3)

Solve for F.