Answer:
The net torque on the square plate is 2.72 N-m.
Explanation:
Given that,
Side = 0.2 m
Force 
Force 
Force 
We need to calculate the torque due to force F₁
Using formula of torque


Put the value into the formula


We need to calculate the torque due to force F₂
Using formula of torque


Put the value into the formula


We need to calculate the torque due to force F₃
Using formula of torque


Put the value into the formula


We need to calculate the net torque on the square plate



Hence, The net torque on the square plate is 2.72 N-m.
We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
Answer:
Fy = 14.3 [N]
Explanation:
To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:
When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

where:
F = 15 [N]
Fx = horizontal component = 4.5 [N]
Fy = vertical component [N]
![15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]](https://tex.z-dn.net/?f=15%3D%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%5C%5C%2015%5E%7B2%7D%3D%20%28%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%29%5E%7B2%7D%20%5C%5C225%20%3D%204.5%5E%7B2%7D%2BF_%7By%7D%20%5E%7B2%7D%5C%5C%20%20F_%7By%7D%5E%7B2%7D%20%3D225%20-4.5%5E%7B2%7D%5C%5C%20F_%7By%7D%5E%7B2%7D%3D204.75%5C%5CF_%7By%7D%3D%5Csqrt%7B204.75%7D%5C%5C%20%20F_%7By%7D%3D14.3%20%5BN%5D)
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