The two forces F ? 1 and F ? 2 (looking down) act on a 28.0?kg object on a frict

Question

The two forces F? 1 and F? 2 (looking down) act on a 28.0?kg object on a frictionless tabletop.

1)   If F1=11.6N and F2=18.2N, find the absolute value of the net force on the object for (a).

2) If

F1=11.6N and F2=18.2N, find the angle of the net force on the object for (a).

3) If

F1=11.6N and F2=18.2N, find the acceleration of the object for (a).

4) If

F1=11.6N and F2=18.2N, find the absolute value of the net force on the object for (b).

5) If

F1=11.6N and F2=18.2N, find the angle of the net force on the object for (b).

6)

Explanation / Answer

Here ,

for perpendicular forces ,

Fnet = sqrt(F1^2 + F2^2)

1)

F1 = 11.6 N

F2 = 18.2 N

Now, as the forces in perpendicular ,

Fnet = sqrt(11.6^2 + 18.2^2)

Fnet = 21.6 N

the net force is 21.6 N

2)

angle = 180 + arctan(F2 / F1)

angle = 180 + arctan(18.2/11.6)

angle = 237.5 degree

the angle of net force is 237.5 degree

3)

using second law of motion ,

Fnet = m*a

a = 21.6/28

a = 0.771 m/s^2

the acceleration is 0.771 m/s^2

4)

for theta = 120 degree ,

Fnet = sqrt(F1^2 + F2^2 + 2* F1 * F2 * cos(theta))

Fnet = sqrt(11.6^2 + 18.2^2 + 2 * 11.6 * 18.2 * cos(120))

Fnet = 16 N

the net force is 16 N

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