A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio
1 answer:
The angle of the wedge is 30°.
Answer:
5.88 ft/s
Explanation:
a) The block will slide down due to it's weight.
initial velocity u= 0
final velocity, v
acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²
Sliding displacement, s = 3ft
Use third equation of motion:
substitute the values and solve for v
b) Use conservation of momentum:
Initial momentum of the system = 0
final momentum = (15) ( 9.8)+ (25)(v')
v' = 5.88 ft/s
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We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:
Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:
Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:
Now, we divide by the distance of the muscle:
Now, we substitute the values:
Now, we solve the operations:
Therefore, the force exerted by the muscles is 23.53 Newtons.
Part B. To determine the force on the pivot we will add the forces we add the vertical forces:
Since there is no vertical movement the sum of vertical forces is zero:
Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:
Now, we plug in the values:
Solving the operations:
Therefore, the force is 73.53 Newtons.
Answer: Answer down below.
Explanation:
Answer:
P = 33.6 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = forces [N]
m = mass = 14 [kg]
a = acceleration = 6 [m/s²]
In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.
where:
W = work [J]
F = force = 84 [N]
d = displaciment = 40 [m]
Finally, the power can be calculated by the relationship between the work performed in a given time interval.
where:
P = power [W]
W = work = 3360 [J]
t = time = 100 [s]
Now replacing:
The power is given in watts