Answer:magnitude -5; angle 160°
Explanation:
Vector A is described as having magnitude 5 and angle -20°.
To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.
5 @ -20° = 5 @ 340°
5 @ -20° = -5 @ 160°
The third one is the answer.
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
Answer:
6495.19 Joule
Explanation:
F = Weight of the crate = 250 N
d = Distance the cart is pushed = 30 m
θ = Angle of inclination = 60°
The weight of the crate will be resloved into two components
Fdsinθ and Fdcosθ
Work done by the force of gravity is
W = Fdsinθ
⇒W = 250×30×sin60
⇒W = 6495.19 Joule
∴ The work done by the force of gravity is 6495.19 Joule
