Math  /  Algebra

Question3. A boy pulls a 47.5 kg crate with a rope. The rope makes an angle of 2828^{\circ} to the horizontal. The coefficient of friction for the crate and the deck is 0.30 . The boy exerts a force of 185 N . What is the acceleration of the crate? (1.04 m/ s2\mathrm{s}^{2} [forward])

Studdy Solution

STEP 1

What is this asking? How fast is a crate speeding up when a boy pulls it with a rope at an angle, considering friction? Watch out! Don't forget to break the force into components and consider *both* the vertical and horizontal components of the pulling force when calculating friction and net force!

STEP 2

1. Calculate the horizontal and vertical components of the pulling force.
2. Calculate the normal force.
3. Calculate the force of friction.
4. Calculate the net force.
5. Calculate the acceleration.

STEP 3

Alright, let's **break down** that pulling force!
We've got the boy pulling at an angle, so some of that force is pulling the crate *forward* and some of it is pulling *upward*.
We need to figure out how much force is doing each!

STEP 4

The **horizontal component** FxF_x is given by Fx=Fcos(θ)F_x = F \cdot \cos(\theta), where FF is the **total pulling force** (185185 N) and θ\theta is the **angle** (2828^\circ).
So, Fx=185 Ncos(28)163.1 NF_x = 185 \text{ N} \cdot \cos(28^\circ) \approx 163.1 \text{ N}.
This is the part of the force that's actually moving the crate forward!

STEP 5

The **vertical component** FyF_y is given by Fy=Fsin(θ)F_y = F \cdot \sin(\theta), where FF is the **total pulling force** (185185 N) and θ\theta is the **angle** (2828^\circ).
So, Fy=185 Nsin(28)86.6 NF_y = 185 \text{ N} \cdot \sin(28^\circ) \approx 86.6 \text{ N}.
This part of the force is *reducing* the force of the crate on the ground!

STEP 6

The **normal force** is the force the ground exerts *upward* on the crate.
Usually, it's just equal to the weight of the crate, but since the boy is pulling *upward* a bit, the normal force is a little less.

STEP 7

The **weight** of the crate is W=mgW = m \cdot g, where mm is the **mass** (47.547.5 kg) and gg is the **acceleration due to gravity** (9.89.8 m/s2^2).
So, W=47.5 kg9.8 m/s2=465.5 NW = 47.5 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 465.5 \text{ N}.

STEP 8

Now, the **normal force** FNF_N is the weight *minus* the vertical component of the pulling force: FN=WFy=465.5 N86.6 N=378.9 NF_N = W - F_y = 465.5 \text{ N} - 86.6 \text{ N} = 378.9 \text{ N}.

STEP 9

Friction! It's always there to slow things down.
The **force of friction** FfF_f is given by Ff=μFNF_f = \mu \cdot F_N, where μ\mu is the **coefficient of friction** (0.300.30) and FNF_N is the **normal force** we just calculated.

STEP 10

So, Ff=0.30378.9 N113.7 NF_f = 0.30 \cdot 378.9 \text{ N} \approx 113.7 \text{ N}.
That's how much force is working *against* the boy's pulling.

STEP 11

The **net force** is the *total* force acting on the crate horizontally.
It's the horizontal component of the pulling force *minus* the force of friction.

STEP 12

So, Fnet=FxFf=163.1 N113.7 N=49.4 NF_{\text{net}} = F_x - F_f = 163.1 \text{ N} - 113.7 \text{ N} = 49.4 \text{ N}.
This is the force that's actually accelerating the crate!

STEP 13

Finally, we can find the **acceleration**!
We use Newton's second law: Fnet=maF_{\text{net}} = m \cdot a, where FnetF_{\text{net}} is the **net force**, mm is the **mass**, and aa is the **acceleration**.

STEP 14

We want to find aa, so we rearrange the formula: a=Fnetma = \frac{F_{\text{net}}}{m}.
Plugging in our values, we get a=49.4 N47.5 kg1.04 m/s2a = \frac{49.4 \text{ N}}{47.5 \text{ kg}} \approx 1.04 \text{ m/s}^2.

STEP 15

The crate accelerates at approximately 1.041.04 m/s2^2 forward.

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