A 4113 N piano is to be pushed up a(n) 3.99 mfrictionless plank that makes an angle of 25.5â—¦with the horizontal.Calculate the
1 answer:
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Answer:
a. Fnet =37.67N
b. The direction = 133.4 from the x axis counter clockwise.
c. Option 2
Explanation:
Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.
Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..
Given that F1 is 71N at 20°, then it is in the first quadrant.
a. Fnet= F1+F2+F3
Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j
Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j
Resolving the vectors into x and y components.
Fnet= -2.04i+37.62j
Magnitude of the vector
Fnet= √((-2.04)^2+(37.62)^2)
Fnet= 37.67N
Fnet is approximately 38N.
b. Direction of the Fnet.
Angle=arctan(y/x)
Angle=arctan(-37.61/2.04)
Angle= -43.37°
The angle is in the negative x axis and positive y axis.
Then the direction becomes 180-43.37
Therefore, the direction of the net force is 133.37°.
c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.
Fnet=ma
Fnet= mv/t
So the velocity is in the direction of the Fnet.
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
