Physics, 11.03.2020 02:26 brookeleneewick

Two forces, F1 vec and F2 vec, act at a point. F1 vec has a magnitude of 9.60 N and is directed at an angle of 55.0∘ above the negative x axis in the second quadrant. F2 vec has a magnitude of 5.40 N and is directed at an angle of 53.1∘ below the negative x axis in the third quadrant. A. What is the x component of the resultant force?B. What is the y component of the resultant force?C. What is the magnitude of the resultant force?

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Answer from: kkstone

A. F_x=-8.74N

B. F_y=3.55N

C. F=9.43N

Explanation:

First, we break down the two vectors F₁ and F₂ into their components. In order to do that, we have to use trigonometry:

$F_{1x}=F_1\cos\theta_1\implies F_{1x}=-9.60N\cos55.0\°=-5.50N\\\\F_{1y}=F_1\sin\theta_1\implies F_{1y}=9.60N\sin55.0\°=7.86N\\\\\\F_{2x}=F_2\cos\theta_2\implies F_{2x}=-5.40N\cos53.1\°=-3.24N\\\\F_{2y}=F_2\sin\theta_2\implies F_{2y}=-5.40N\sin53.1\°=-4.31N\\\\$

(Note: $F_{1x}$ , $F_{2x}$ and $F_{2y}$ are negative because the vectors are in the second and third quadrant, respectively. So those components must be negative)

Next, we sum the x components and the y components to get the components of the resultant force:

$F_x=F_{1x}+F_{2x}=-5.50N-3.24N=-8.74N\\\\F_y=F_{1y}+F_{2y}=7.86N-4.31N=3.55N$

So, the x component of the resultant force is -8.74N (A), and its y component is 3.55N (B).

Finally, to get the magnitude of the resultant force, we have to use the Pythagorean Theorem:

$F=\sqrt{F_x^{2}+F_y^{2}}\\ \\F=\sqrt{(-8.74N)^{2}+(3.55N)^{2}}=9.43N$

In words, the magnitude of the resultant force is 9.43N (C).

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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