A horse is harnessed to a sled having a mass of 236 kg, includ- ing supplies. The horse must exe

step 1 So we're told we have a horse that's harnessed to a sled. The sled has a mass of 236 kilograms.

A horse is harnessed to a sled having a mass of 236 kg,...

A horse is harnessed to a sled having a mass of $236 \mathrm{kg},$ includ- ing supplies. The horse must exert a force exceeding 1240 $\mathrm{N}$ at an angle of $35.0^{\circ}$ in order to get the sled moving. Treat the sled as a point particle. (a) Calculate the normal force on the sled when the magnitude of the applied force is 1 240 N. (b) Find the coefficient of static friction between the sled and the ground beneath it. (c) Find the static friction force when the horse is exerting a force of $6.20 \times 10^{2} \mathrm{N}$ on the sled at the same angle.

Key Concepts

Free-Body Diagrams

A free-body diagram is a simplified representation of all forces acting on an object. It involves sketching the object and drawing vectors for gravitational force, normal force, applied forces, and friction. This method helps visualize how forces interact and is crucial for setting up the equilibrium equations used to solve mechanics problems.

Decomposing Forces

Decomposing forces means breaking a force vector into its perpendicular components, typically horizontal and vertical. This is done using trigonometric functions such as sine and cosine, and it is essential when analyzing problems where forces are applied at angles. It enables the separation of the force into parts that affect motion in different directions.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object. It typically counteracts the object's weight and any additional vertical forces. Calculating the normal force is vital because it influences other forces, such as friction, which depend on the magnitude of the normal force.

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It adjusts up to a maximum value to prevent relative motion. Understanding static friction is critical to determine how much force is needed to start moving an object, particularly when additional forces modify the effective normal force.

Coefficient of Friction

The coefficient of friction is a dimensionless constant that characterizes the interaction between two surfaces. Specifically, the coefficient of static friction represents the ratio between the maximum static frictional force and the normal force. It is a key parameter when calculating the threshold force required to overcome static friction and initiate movement.

Transcript

00:01 So we're told we have a horse that's harnessed to a sled.

00:04 The sled has a mass of 236 kilograms.

00:09 The horse must exert a force of 1 ,400, 1 ,240 newtons at an angle of 35 degrees with the horizontal to get the sled moving.

00:21 So we're going to treat the sled as a point particle.

00:24 So again, all these forces basically act through this point here.

00:28 We want to calculate the normal force on the sled when the magnitude of the side.

00:31 The force of the applied force is this value here.

00:37 Well, we can look at forces in the y direction.

00:43 Okay, so we have, let me put a little axes in here.

00:46 We have y and have x.

00:53 So a summation of forces in the y direction is going to be equal to zero because it's not accelerating in that direction.

00:59 So we have n plus f sign of 35 degrees, my minus w equals zero.

01:08 So n equals the weight, mg, minus f sine of 35 degrees, and that winds up being 1 ,600 newtons.

01:19 So you can see here that the force is actually, the horse is actually lifting this off the ground so that the normal force is actually a little bit less, that is less than the weight because the horse is off the, part of the force that the horse is applying is also lifting acting to lift the sled.

01:42 So let's see here.

01:45 That's the normal force.

01:47 So then we want to calculate the coefficient of static friction between the sled and the ground beneath it.

01:51 All right.

01:51 So that means we're going to look at the forces in the x direction.

01:55 And that means we have minus the friction force plus f cosine of 35 degrees.

02:03 And again, we're assuming that it hasn't started accelerating...