We read your photo as:

A garden roller with a mass of 200 kg is pushed by a man at a constant speed over a lawn by exerting a constant force of 400 N on the handle of the roller that forms an angle of 60° to the horizontal.

Physics

ScanSolve AI Answer

Answers

4 parts
7.1

200 N\text{200 N}

7.2

346.41 N\text{346.41 N}

7.3

1960 N\text{1960 N}

7.4

2306.41 N\text{2306.41 N}

Step-by-step solution

4 steps

Here are the solutions to the problems:

7. A garden roller with a mass of 200 kg is pushed by a man at a constant speed over a lawn by exerting a constant force of 400 N on the handle of the roller that forms an angle of 60° to the horizontal.

Given:
Mass of the roller, m=200m = 200 kg
Applied force, F=400F = 400 N
Angle below the horizontal, θ=60°\theta = 60°
Acceleration due to gravity, g=9.8 m/s2g = 9.8 \text{ m/s}^2

Part 7.1· 1 step

Calculate the force that causes the roller to move forward.

This is the horizontal component of the applied force.

  1. Calculate the horizontal component (FxF_x).

    Fx=Fcos⁡θF_x = F \cos \theta Fx=400 N×cos⁡(60°)F_x = 400 \text{ N} \times \cos(60°) Fx=400 N×0.5F_x = 400 \text{ N} \times 0.5 Fx=200 NF_x = 200 \text{ N}

    The force that causes the roller to move forward is 200 N\boxed{\text{200 N}}.

Part 7.2

346.41 N\text{346.41 N}

Calculate the force that tends to pin the roller to the ground.

This is the vertical component of the applied force, directed downwards.

  1. Calculate the vertical component (FyF_y).

    Fy=Fsin⁡θF_y = F \sin \theta Fy=400 N×sin⁡(60°)F_y = 400 \text{ N} \times \sin(60°) Fy=400 N×32F_y = 400 \text{ N} \times \frac{\sqrt{3}}{2} Fy≈400 N×0.8660F_y \approx 400 \text{ N} \times 0.8660 Fy≈346.41 NF_y \approx 346.41 \text{ N}

    The force that tends to pin the roller to the ground is 346.41 N\boxed{\text{346.41 N}}.

Part 7.3

1960 N\text{1960 N}

Calculate the weight (FgF_g) of the roller.

  1. Calculate the weight using the formula Fg=m×gF_g = m \times g.

    Fg=m×gF_g = m \times g Fg=200 kg×9.8 m/s2F_g = 200 \text{ kg} \times 9.8 \text{ m/s}^2 Fg=1960 NF_g = 1960 \text{ N}

    The weight of the roller is 1960 N\boxed{\text{1960 N}}.

Part 7.4

2306.41 N\text{2306.41 N}

Calculate the force that is exerted by the roller on the lawn, i.e., the effective weight of the roller when it is being pushed.

The effective weight is the sum of the roller's actual weight and the downward vertical component of the applied force.

  1. Add the weight and the vertical component.

    Effective weight=Fg+Fy\text{Effective weight} = F_g + F_y Effective weight=1960 N+346.41 N\text{Effective weight} = 1960 \text{ N} + 346.41 \text{ N} Effective weight=2306.41 N\text{Effective weight} = 2306.41 \text{ N}

    The effective weight of the roller when it is being pushed is 2306.41 N\boxed{\text{2306.41 N}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

Q7.6 How does the magnitude and direction of the horizontal force of the lawn that counters the…Q7.5 What is the magnitude and direction of the horizontal force that counters the movement of the…QA force of magnitude 50 N is acting at an angle of 30° on a 2 kg object to push the object against a…QA bodybuilder pulls a 60 kg tyre with a force of 250 N at an angle of 20 degrees to the horizontal…QA man of mass 50 kg stands on a crate of mass 200 kg. The rope attached to the crate passes over a…
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp