1. A horse is harnessed to a sled having a mass of 231 kg, including supplies. The horse must exert a force exceeding 1250 N at an angle of 36.3° (above the horizontal) in order to get the sled moving. Treat the sled as a point particle.
(a)Calculate the normal force (in N) on the sled when the magnitude of the applied force is 1250 N. (Enter the magnitude.)
(b)Find the coefficient of static friction between the sled and the ground beneath it.
(c)Find the static friction force (in N) when the horse is exerting a force of 6.25 ✕ 102 N on the sled at the same angle. (Enter the magnitude.)
2. Consider the figure below.
(a) Find the tension in each cable supporting the 688-N cat burglar. (Assume the angle θ of the inclined cable is 38.0°.)
inclined cable - ???
horizontal cable - ???
vertical cable - ???
www.webassign.net/sercp9/4-p-017-alt.gif
(a) The tension in the inclined cable is 545.2 N, the tension in the horizontal cable is 688 N, and the tension in the vertical cable is 545.2 N.
1. (a) To calculate the normal force on the sled, we need to consider the vertical component of the applied force. The vertical component is given by:
Vertical component = Applied force * sin(angle)
= 1250 N * sin(36.3°)
≈ 736.02 N
The normal force is equal in magnitude but opposite in direction to the vertical component of the applied force. Therefore, the normal force is approximately 736.02 N.
(b) The coefficient of static friction (μ_static) can be calculated using the following equation:
μ_static = (Force of static friction) / (Normal force)
From part (a), the normal force is approximately 736.02 N. Since the sled is not moving, the force of static friction is equal in magnitude but opposite in direction to the horizontal component of the applied force.
Horizontal component = Applied force * cos(angle)
= 1250 N * cos(36.3°)
≈ 1018.20 N
Therefore, the coefficient of static friction is:
μ_static = 1018.20 N / 736.02 N
≈ 1.385
(c) When the horse exerts a force of 6.25 × 10^2 N on the sled at the same angle (36.3°), we can calculate the static friction force using a similar approach as in part (b).
Horizontal component = Applied force * cos(angle)
= 6.25 × 10^2 N * cos(36.3°)
≈ 510.79 N
Therefore, the static friction force is approximately 510.79 N.
2. In order to find the tension in each cable supporting the 688 N cat burglar, we need to break down the forces acting on the cat burglar.
(a) For the inclined cable:
Vertical component of tension = weight of cat burglar * cos(θ)
= 688 N * cos(38.0°)
≈ 552.28 N
Horizontal component of tension = weight of cat burglar * sin(θ)
= 688 N * sin(38.0°)
≈ 420.89 N
Therefore, the tension in the inclined cable is approximately 552.28 N (vertical) and 420.89 N (horizontal).
(b) For the horizontal cable:
The horizontal cable only supports the horizontal component of tension, which is 420.89 N. Therefore, the tension in the horizontal cable is approximately 420.89 N.
(c) For the vertical cable:
The vertical cable only supports the vertical component of tension, which is 688 N. Therefore, the tension in the vertical cable is approximately 688 N.
To solve these types of physics problems, we'll need to use Newton's laws of motion and some trigonometry. Let's break down each question and go through the steps to find the answers.
1. (a) To calculate the normal force on the sled, we'll start by drawing a free-body diagram. The forces acting on the sled are the gravitational force (mg) and the force applied by the horse (F). Since the sled is on a horizontal surface, the normal force (N) will be equal in magnitude but opposite in direction to the gravitational force.
Using Newton's second law (F = ma), we can set up the equation:
N - mg = 0
N = mg
The mass of the sled is given as 231 kg, and acceleration due to gravity is approximately 9.8 m/s^2. Plugging in these values:
N = (231 kg)(9.8 m/s^2)
Calculating this will give us the normal force on the sled.
(b) To find the coefficient of static friction between the sled and the ground, we can use the equation:
μ_s = F_friction / N
Since the sled is just about to start moving, the applied force is equal to the maximum static friction force. Here, the applied force is given as 1250 N. Plug this value into the equation along with the normal force calculated in part (a) to find the coefficient of static friction.
(c) When the horse exerts a force of 6.25 x 10^2 N on the sled at the same angle (36.3°), we need to find the static friction force. We can use trigonometry to split the force into horizontal and vertical components.
The vertical component (F_vertical) is given by F * sin(angle), and the horizontal component (F_horizontal) is given by F * cos(angle). Since the sled is not moving vertically, the static friction force must match the vertical component.
Set up the equation:
F_friction = F_vertical = F * sin(angle)
Plug in the given force (6.25 x 10^2 N) and the angle (36.3°) to calculate the static friction force.
2. To find the tension in each cable supporting the 688-N cat burglar, we'll use the concept of equilibrium. For an object in equilibrium, the sum of all forces acting on it must be zero.
Draw a free-body diagram for the cat burglar. There are three cables supporting the cat burglar: an inclined cable, a horizontal cable, and a vertical cable. The tension in each cable is the force acting away from the cat burglar parallel to the cable.
Using the tension in the inclined cable, the tension in the horizontal cable, and the tension in the vertical cable, set up the equilibrium equations based on the forces acting along the x- and y-axes.
For example, in the x-direction, the tension in the inclined cable can be split into horizontal and vertical components. The horizontal component (T_x) is given by T * cos(angle), where T is the tension in the cable and angle is 38.0°.
Solve similar equations for the forces acting in the y-direction as well, relating them to the weight of the cat burglar.
Once you have the equations set up, solve them simultaneously to find the tension in each cable.
Remember to use the given values of the weight of the cat burglar (688 N) and the angle of the inclined cable (38.0°).