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Rob is pushing a block of 12 kg up a frictionless ramp. The ramp makes an angle of 35Â° with the horizontal. Assuming that Rob exerts a force of 148 N, find:

- The
*resultant force*acting on the block. - The
*acceleration*of the block. - The magnitude of the
*normal force*.

Let's represent what is happening in the problem with a simple sketch.

We need to represent a ramp that makes an angle of 35Â° with the horizontal, a block on it, and indicate that the block is pushed up by a force parallel to the ramp:

By looking at the sketch, we can conclude that there are three forces acting on the block:

- Rob's push, F
- the gravitational force, mg
- and the normal force N, which the ramp exerts to prevent the penetration of the block

Having figured out the forces acting on the block, let's draw a free-body diagram of the block:

Let's think about what we know, and what we're asked to find:

We know the mass of the block (12 kg), the angle the ramp makes with the horizontal (35Â°), and the force exerted by Rob (148 N).

We're asked to *find* the resultant force acting on the block, the acceleration that the block has as a result, and the normal force exerted by the ramp on the block.

m = 12 kg

Î¸ = 35Â°

F = 148 N

R = ?

a = ?

N = ?

Let's start by finding the resultant force.

In this case we have a block moving up a ramp, so for our convenience, we will use tilted coordinate axes, with the x axis in the direction of motion (uphill).

After drawing the coordinate axes on the free-body diagram of the block, we proceed to find the components of the individual forces acting on the block:

Keep in mind that the angle that the gravitational force, mg, makes with its y component, mg_{y}, is equal to the angle that the ramp makes with the horizontal (35Â° in this case).

F_{x} = F

N_{x} = 0

mg_{x} = âˆ’mg sin 35Â°

F_{y} = 0

N_{y} = N

mg_{y} = âˆ’mg cos 35Â°

And the components of the resultant force will be:

x:

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

R_{x} = F + 0 + (âˆ’mg sin 35Â°)

R_{x} = F âˆ’ mg sin 35Â° (1)

y:

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

R_{y} = 0 + N + (âˆ’mg cos 35Â°)

R_{y} = N âˆ’ mg cos 35Â° (2)

We know that the motion of the block is along the ramp, therefore R_{y} must be zero. Otherwise there would be an acceleration in the y direction, which is not the case.

R_{y} = 0

Therefore, the magnitude of R is equal to the absolute value of R_{x}. We know that R_{x} must be positive because the block is accelerating in the positive x direction. So, we can write:

R = R_{x}

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

R_{x} = F âˆ’ mg sin 35Â°

Therefore,

R = F âˆ’ mg sin 35Â°

R = 148 N âˆ’ (12 kg) (9.8 N/kg) (sin 35Â°)

R = 148 N âˆ’ 67 N

R = 81 N

So, the resultant force on the block is 81 N and directed in the positive x direction:

Next we need to find the acceleration.

Since we know the resultant force and the mass, we can apply **Newton's 2 ^{nd} Law** to get the acceleration:

R = ma

a = | R |

m |

a = | R |

m |

a = | 81 N |

12 kg |

a = 6.8 m/s^{2}

Lastly, we need to find the magnitude of the normal force.

The normal force N appears in Eq. (2):

R_{y} = N âˆ’ mg cos 35Â°

And we've shown that R_{y} is zero, therefore:

0 = N âˆ’ mg cos 35Â°

N = mg cos 35Â°

N = (12 kg) (9.8 N/kg) (cos 35Â°)

N = 96 N

At this point, we've found everything we were asked to find:

R = 81 N

a = 6.8 m/s^{2}

N = 96 N

- When an object is upon an incline, the angle between mg and mg
_{y}is equal to the angle that the incline makes with the horizontal. - For objects moving along an incline, the y component of the resultant force is always zero (assuming that we take the x axis in the direction of motion, and the y axis perpendicular to it).

A heavy box of 135 kg is pushed up an inclined plane which makes an angle of 14.0Â° with the horizontal.

If the push has magnitude 550 N and the plane can be considered frictionless, what are: the resultant force acting on the box, the acceleration of the box, and the normal force?

R = 230 N

a = 1.70 m/s^{2}

N = 1.29 Ã— 10^{3} N

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