Two forces (35.N at 75\" north of east, and 48.N at 39\" south of west) are forc

Question

Two forces (35.N at 75" north of east, and 48.N at 39" south of west) are force is determined by these two forces, then fill in the following values. Draw g ap your work leading to your answers. Assume SI units are being used. Find the work done by this The force acting on a particle is F_x- 2x^2 + 10x + 3. Assume SI units are being used. Find the work bone by this force from x = 4_m to x = 8_m. A rod with uniform mass distribution (12 kg, 2 m, axis through the center) is to be accelerated from 5 rad/s to 127 rad/s in just 3 s. Find the net torque that must be applied to the rod to accomplish this. A rocket is launched and attains a speed of 4900_m/s when 49000.m above the surface of the Earth. If the engines then shut down, how much higher will the rocket rise before beginning its fall back toward the Earth

Explanation / Answer

First we have to calculate the net force in X and Y direction.
So we will resolve the forces in x and y direction
35 N at an angle of 75 degree therefore
Along X = 35 Cos 75 = 9.0567 N
Along Y = 35Sin35 = 20.075 N
Similarly the next force is 48 N at an angle of 39 degree
So along X axis = -48Cos39 = - 37.3 N (minus sign shows that the direction is opposite or toward negative X direction)
Along Y axis
= -48Sin39 = -30.207 N
Therfore
(27) along X = 9.05867 - 37.3 = -28.241 N
Along Y = 33.8 - 30.2 = 3.6 N
(28) Now in x direction
-28.241 = max
aX =- 0.0941 m/s2 (minus sign shows that the acceleration is toward -X direction)
similarly in Y direction
3.6 = maY
aY = 0.012 m/s2
(29) a = (ax2  +aY2)1/2 = 0.09486 m/s2
tan(theta) = aY/aX
Theta = -7.26 degree
that is the angle from the - X direction

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