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A crate is pulled to the right with a force of 85 N, to the left with a force of 115 N, upward with a force of 565 N, and downward with a force of 236 N.

1. Find the magnitude and direction of the net external force on the crate.

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Answer:

330.4 N and  [tex]95.2^{\circ}[/tex]counterclocwise to the x direction

Explanation:

Sum of forces in vertical are equal, let movement to right and upwards be positive while left and downwards be negative

Net force in horizontal direction is 85-115=-30 N

Net force in vertical direction is 565-236=329 N

Resultant force=[tex]\sqrt {(-30)^{2}+329^{2}}=330.3649497 N\approx 330.4 N[/tex]

Direction=[tex]tan^{-1}\frac {329}{-30}=-84.78986932\approx -84.8^{\circ}[/tex]

180-84.8=95.2

Therefore, direction is [tex]95.2^{\circ}[/tex]counterclocwise to the x direction

The magnitude and direction of the net external force on the crate is;

Magnitude; R = 333.36 N

Direction; 95.21° in the counterclockwise direction from the positive x-axis.

We are given;

Force to the right; F_r = 85N

Force to the left; F_l = -115 N (It is negative because on x and y coordinates, left of x-axis denotes approaching negative section)

Force upwards; F_up = 565 N

Force downwards; F_dwn = -236 ((It is negative because on x and y coordinates, down of y-axis denotes approaching negative section)

Now, let us find the net force in the horizontal direction;

ΣF_x = 85 - 115

ΣF_x = -30 N

Similarly, for the vertical direction, we have;

ΣF_y = 565 - 236

ΣF_y = 329 N

The magnitude of the resultant will be;

R = √[(ΣF_x)² + (ΣF_y)²]

R = √[(-30)² + (329)²]

R = √109141

R = 333.36 N

The direction of the resultant is;

θ = tan⁻¹(ΣF_y/ΣF_x)

θ = tan⁻¹(329/-30)

θ = -84.79°

Since angle is negative, it means that it is in the counterclockwise direction from the positive x-axis.

Thus, the direction will be;

θ = -84.79° + 180°

θ = 95.21°

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