Bif and Hap both want the 12.0 kg file cabinet. Bif pulls to the left with a force of 200 N. Hap exerts a force of 200 N to the right, at an angle of 30 degree above horizontal.
a.)Draw the free-body diagram of the cabinet with all forces labeled and calculated.
b.)What is the speed and direction of the cabinet 2.0 s after they both start pulling?
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a) Free-body diagram:
In this scenario, we have two forces acting on the file cabinet - one to the left exerted by Bif and the other to the right exerted by Hap at an angle of 30 degrees above horizontal.
The free-body diagram would look like this:
------------
| F(Hap)
| (30 )
|
------------
File
Cabinet
---------
| |
| |
---------
The force exerted by Bif is 200 N to the left and the force exerted by Hap is also 200 N but to the right at an angle of 30 degrees above the horizontal.
b) Speed and direction of the cabinet 2.0 s after they start pulling:
To calculate the speed and direction, we need to consider the Net Force and apply Newton's second law, F = ma.
Net Force (F_net) is the vector sum of the forces acting on the cabinet.
F_net = F_Hap - F_Bif
F_net = 200 N (cos 30°) - 200 N
F_net = (200 N * 0.866) - 200 N
F_net = 173.21 N - 200 N
F_net = -26.79 N
The negative sign indicates that the net force is in the opposite direction of Bif's force.
Using Newton's second law, F = ma, we can find the acceleration of the cabinet.
F_net = ma
-26.79 N = (12 kg) * a
a = -26.79 N / 12 kg
a ≈ -2.23 m/s²
The negative sign indicates that the acceleration is in the opposite direction of Bif's force.
To find the speed of the cabinet after 2 seconds, we can use the formula:
v = u + at
Since the initial velocity (u) is 0 m/s (assuming the cabinet starts from rest), the equation simplifies to:
v = at
v = -2.23 m/s² * 2 s
v = -4.46 m/s
The negative sign indicates that the direction of the cabinet is opposite to the direction of Bif's force. So, the cabinet is moving to the left with a speed of 4.46 m/s after 2.0 seconds.
a.) To draw the free-body diagram of the cabinet, we need to consider all the forces acting on it.
First, we have Bif pulling to the left with a force of 200 N. This force can be drawn as an arrow pointing to the left.
Next, we have Hap exerting a force to the right at an angle of 30 degrees above the horizontal. To determine the horizontal and vertical components of this force, we can use trigonometry. The horizontal component (F_horizontal) can be found by multiplying the total force (200 N) by the cosine of the angle (30 degrees). The vertical component (F_vertical) can be found by multiplying the total force (200 N) by the sine of the angle (30 degrees).
Once we have the horizontal and vertical components, we can draw them as arrows. The horizontal component will be pointing to the right, while the vertical component will be pointing up.
Finally, we have the weight of the file cabinet acting downward. The weight can be calculated using the formula W = m * g, where m is the mass of the cabinet (12.0 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). The weight can be drawn as an arrow pointing downward.
b.) To calculate the speed and direction of the cabinet 2.0 s after they both start pulling, we need to use Newton's second law of motion. The net force acting on the cabinet will be the sum of all the forces. We can find the net force by subtracting the force exerted by Bif (200 N) from the force exerted by Hap (F_horizontal).
The net force can be calculated as:
Net force = F_horizontal - Force by Bif
Net force = (200 N * cos(30°)) - 200 N
Once we have the net force, we can use Newton's second law: F = m * a, where F is the net force, m is the mass of the cabinet, and a is the acceleration of the cabinet.
Since the force is acting horizontally, the acceleration will also be horizontal. We can calculate the acceleration using the formula a = F / m.
After finding the acceleration, we can use the equation for uniformly accelerated motion to find the speed at a given time (2.0 s) after they start pulling. The equation is v = u + at, where v is the final velocity, u is the initial velocity (which we assume to be 0 m/s since the cabinet is initially at rest), a is the acceleration, and t is the time.
Thus, by substituting the values into the equation, we can find the speed and direction of the cabinet 2.0 s after they start pulling.
F1 = -200 N.
F2 = 200N.[30o].
Fr = F1 + F2 = -200 + 200[30].
Fr = (-200+200*Cos30) + (200*sin30)I = -26.8 + 100i = 103.5N.[-75o].
Fr = 103.5N.[75o] N. of W. = 103.5N.[105o] CCW from +x-axis.
Fr = Resultant force.
b. Fr = M*a.
a = Fr/M = 103.5[105]/12 = 8.625 m/s^2[105o].
V = Vo + a*T = 0 + 8.625[105o]*2 = 17.25 m/s[105o]