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- â€º This problem

A box of 5.0 kg stays on a frictionless horizontal surface.

Bob then starts pulling the box with a force of 100 N at an angle of 15Â° with the horizontal.

- Draw a
*free-body diagram*for the box. - What is the
*resultant force*acting on the box? - What is the
*acceleration*of the box? - What is the
*normal force*that the surface exerts on the box?

Before we can solve this problem, we need to understand what is happening. Let's try to draw a sketch, using the information that was specified in the problem description.

So, we probably need to draw a horizontal surface, a rectangular box sitting on the surface, and indicate that the box is being pulled by a force that makes 15Â° with the horizontal:

The next step is to look carefully at our sketch, and try to enumerate all the forces that are acting on the box.

There are clearly 3 forces acting:

- Bob's pulling force, which we'll indicate with F
- the gravitational force mg
- and the normal force N, which the surface exerts to prevent the penetration of the box

At this point we are ready to draw a free-body diagram for the box:

Now, let's carefully look at what we know, and what the problem is asking:

We *know* the mass of the box (5.0 kg) and Bob's pull (100 N at 15Â° angle with the horizontal).

The problem is *asking* us to draw a free-body diagram for the box (which we already did), then to determine the resultant force that acts on the box, the acceleration that the box has as a result, and the normal force that the surface exerts on the box.

In short:

m = 5.0 kg

F = 100 N

Î¸ = 15Â°

R = ?

a = ?

N = ?

We can start by finding the resultant force.

So, let's draw the coordinate axes on our diagram, with the x axis in the direction of motion, and find the x and y components of the 3 forces that act on the box:

F_{x} = F cos 15Â°

N_{x} = 0

mg_{x} = 0

F_{y} = F sin 15Â°

N_{y} = N

mg_{y} = âˆ’mg

We can now find the x and y components of the resultant force:

x:

R_{x} = F_{x} + N_{x} + mg_{x}

R_{x} = F cos 15Â° + 0 + 0

R_{x} = F cos 15Â° (1)

y:

R_{y} = F_{y} + N_{y} + mg_{y}

R_{y} = F sin 15Â° + N + (âˆ’mg)

R_{y} = F sin 15Â° + N âˆ’ mg (2)

Since the motion is along a horizontal surface, the y component of the resultant force, R_{y}, must be zero. Otherwise there would be an acceleration in the y direction (for **Newton's 2 ^{nd} Law**), which is not the case.

R_{y} = 0

Therefore, the magnitude of R will be equal to the magnitude of its x component.

The x component, R_{x}, is positive because the box is accelerating in the positive x direction:

R_{x} > 0

R = R_{x}

We have already determined R_{x} in Eq. (1):

R_{x} = F cos 15Â°

R = R_{x} = F cos 15Â°

R = (100 N) (cos 15Â°)

R = 97 N

And R will have direction in the positive x axis:

Next, let's find the acceleration of the box.

We know the mass of the box, and we just found the resultant force acting on the box. We can apply **Newton's 2 ^{nd} Law** to get the acceleration:

R = ma

a = | R |

m |

The direction of a is the same as that of R. Let's calculate the magnitude of a:

a = | R |

m |

a = | 97 N |

5.0 kg |

a = 19 N/kg = 19 m/s^{2}

The last unknown on our list is the normal force N.

N has direction in the positive y axis. But what about the magnitude?

We saw N appear in Eq. (2):

R_{y} = F sin 15Â° + N âˆ’ mg

And we know that R_{y} is zero, therefore:

0 = F sin 15Â° + N âˆ’ mg

This is an equation with one unknown (N). So, let's solve it:

0 = F sin 15Â° + N âˆ’ mg

âˆ’N = F sin 15Â° âˆ’ mg

N = âˆ’F sin 15Â° + mg

N = mg âˆ’ F sin 15Â°

N = (5.0 kg) (9.8 N/kg) âˆ’ (100 N) (sin 15Â°)

N = 49 N âˆ’ 26 N

N = 23 N

We have found all the unknowns the problem asked us:

R = 97 N

a = 19 m/s^{2}

N = 23 N

- Don't make the mistake to assume that the magnitude of the normal force is
*always equal*to the magnitude of the gravitational force mg. As we saw in this problem, the normal force had magnitude equal to mg âˆ’ F sin 15Â°. - Since the definition of
**newton**is: 1 N = 1 kg Â· m/s^{2}, it follows that: 1 N/kg = 1 m/s^{2}. This means that we can use N/kg and m/s^{2}interchangeably to express accelerations. - If an object moves horizontally, the vertical component of the resultant force is zero.

A crate of mass 31.0 kg is pulled over the floor with a force of 275 N. The force makes an angle of 24.0Â° with the horizontal. Assuming there is no friction between the crate and the floor, find the resultant force acting on the crate, the acceleration of the crate, and the magnitude of the normal force.

R = 251 N

a = 8.10 m/s^{2}

N = 192 N

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