Assume the sleigh and horse pulling it are on a level, horizontal surface.
Given: mass of sleigh 475 kg / force of horse pulling sleigh 1.5×10^5 / coef. of kinetic friction .15 / coef. of static friction .20
a.) What is the normal force of the sleigh and its passengers?
b.) What is the maximum static friction force needed to get the sleigh to move?
c.) Once the sleigh is moving, what is the force of kinetic friction?
d.) What is the magnitude and direction of the net force when the sleigh is moving?
e.) What is the magnitude and direction of the acceleration of the sleigh?
a. M*g = 475 * 9.8 = 4655 N. = Wt. of sleigh. = Normal force(Fn).
b. Fs = u*Fn = 0.20 * 4655 =
c. Fk = uk*Fn = 0.15 * 4655 =
d. Fnet = Fh-Fk = 150,000-4655 =
e. Fnet = M*a.
a = Fnet/M.
To solve these problems, we will use Newton's laws of motion and the equations related to friction. Here's how we can find the answers to each question:
a.) The normal force (N) of the sleigh and its passengers can be found using the equation N = mg, where m is the mass of the sleigh and g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, N = (475 kg)(9.8 m/s^2) = 4655 N.
b.) To find the maximum static friction force needed to get the sleigh to move, we can use the equation fs(max) = μsN, where μs is the coefficient of static friction. Using the given coefficient of static friction (μs = 0.20) and the normal force (N = 4655 N) that we found from part (a), fs(max) = (0.20)(4655 N) = 931 N.
c.) Once the sleigh is moving, the force of kinetic friction (fk) can be found using the equation fk = μkN, where μk is the coefficient of kinetic friction. Given the coefficient of kinetic friction (μk = 0.15) and the normal force (N = 4655 N) from part (a), fk = (0.15)(4655 N) = 698.25 N.
d.) When the sleigh is moving, the net force is the difference between the force applied by the horse and the force of kinetic friction. The force applied by the horse is 1.5×10^5 N (given) and the force of kinetic friction is 698.25 N (from part c). Therefore, the net force is 1.5×10^5 N - 698.25 N = 1.49301×10^5 N in the direction of the horse's pull.
e.) The acceleration of the sleigh can be found using Newton's second law of motion, F = ma, where F is the net force and a is the acceleration. We already found the net force (1.49301×10^5 N) in part (d), and we know the mass of the sleigh (475 kg). Therefore, the acceleration (a) = F/m = (1.49301×10^5 N)/(475 kg) = 313.90 m/s^2 in the direction of the horse's pull.
In summary:
a.) The normal force of the sleigh and its passengers is 4655 N.
b.) The maximum static friction force needed to get the sleigh to move is 931 N.
c.) Once the sleigh is moving, the force of kinetic friction is 698.25 N.
d.) The magnitude and direction of the net force when the sleigh is moving is 1.49301×10^5 N in the direction of the horse's pull.
e.) The magnitude and direction of the acceleration of the sleigh is 313.90 m/s^2 in the direction of the horse's pull.