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Answer:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N
Answer:
d. 3332.5 [N]
Explanation:
To solve this problem we will use newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.
Here we have two forces, the force that pushes the car to move forward and the friction force.
The friction force is equal to the product of the normal force by the coefficient of friction.
f = N * μ
f = (m*g) * μ
where:
N = weight of the car = 2150*9.81 = 21091.5 [N]
μ = 0.25
f = (21091.5) * 0.25
f = 5273 [N]
Now as the car is moving forward, the car wheels move clockwise. The friction force between the wheels of the car and the pavement must be counterclockwise, i.e. counterclockwise. Therefore the direction of this force is forward. This way we have:
F + f = m*a
F + 5273 = 2150*4
F = 8600 - 5273
F = 3327 [N]
Therefore the answer is d.
