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Basile [38]
3 years ago
6

A heavy box weighing 1000 newtons sits on the floor. You lift upward on the box with a force of 450 newtons, but the box does no

t move. What is the normal force on the box while you are lifting?
Physics
2 answers:
Svetllana [295]3 years ago
5 0
The answer for sure is 550

Ex: 1000 is the box
450 is your force
1000-450=550
Anvisha [2.4K]3 years ago
3 0

Answer:

i think its 550

Explanation:

1000-450 is 550

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A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .
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A) Net force in component form: F=100.0i-300.0j

B) Magnitude of the net force: 316.2, direction: -71.6^{\circ}

Explanation:

A)

The three forces given in this problem are:

F_1=300i+500j

F_2=-200i

F_3=-800 j

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100

- Along the y-direction:

F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300

So, the net force in component form is

F=100.0i-300.0j

B)

The magnitude of a vector F is given by Pythagorean's theorem:

|F|=\sqrt{F_x^2+F_y^2}

where in this problem,

F_x=100 is the x-component

F_y=-300 is the y-component

Substituting,

|F|=\sqrt{(100)^2+(-300)^2}=316.2

The direction instead is given by

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

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