A force of 21.0 n is required to start a 2.8 kg box moving across a horizontal concrete floor. (a) what is the coefficient of st
1 answer:
(a)
Assuming the box is on a horizontal surface, the Normal Force would be equal to the Gravitational force which is 2.8 * 9.8 = 27.44N. You are already given the force, so we are ready to solve:
Since a coefficient of friction doesn't have units, your answer is 0.77.
(b)
We can use F = ma to get the Force, and then the same Normal force to get the kinetic coefficient of friction.
We can use that to formulate the kinetic coefficient of friction:
And those are your answers. Hope this helps. I'm not a college student, but I'm taking AP Physics, so I have an idea.
Assuming the box is on a horizontal surface, the Normal Force would be equal to the Gravitational force which is 2.8 * 9.8 = 27.44N. You are already given the force, so we are ready to solve:
Since a coefficient of friction doesn't have units, your answer is 0.77.
(b)
We can use F = ma to get the Force, and then the same Normal force to get the kinetic coefficient of friction.
We can use that to formulate the kinetic coefficient of friction:
And those are your answers. Hope this helps. I'm not a college student, but I'm taking AP Physics, so I have an idea.
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Answer:
Explanation:
<u>Uniform Acceleration </u>
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:
Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.
Solving for a:
Substituting:
Answer:
a) 29062.125 N·m
b) 0 N·m
c)
d) T = 11186.02 N
Explanation:
We are given
Beam mass = 1975 kg
Beam length = 3 m
Cable angle = 60° above horizontal
a) We have the formula for torque given as follows;
Torque about the pin = Force × Perpendicular distance of force from pin
Where the force = Force due to gravity or weight, we have
Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N
Point of action of force = Midpoint for a uniform beam = length/2
∴ Point of action of force = 3/2 = 1.5 m
Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m
b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;
Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m
c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows
Sum of moments about p is given as follows
∑M = 0 gives;
T·sin(θ) × L= M×L/2×g
Therefore torque due to tension is given by the following expression
d) Plugging in the values in the torque due to tension equation, we have;
Therefore, we make the tension force, T the subject of the formula hence
Answer:
Solution:
As per the question:
Mass of first object, m = 120 kg
Mass of second object, m' = 420 kg
Mass of the third object, M = 69.0 kg
Distance between the m and m', d = 0.380 m
Now,
To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:
To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':
To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':