(a) (m * g) - [1250 * cos(36.3º)]
(b) [result from (a)] / [1250 * sin(36.3º)]
(c) [result from (b)] * {(m * g) - [625 * cos(36.3º)]}
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.
HINT - Use the relation between the normal force and the maximum static friction force, solving for μs, the coefficient of static friction.
Apply the x-component of Newton's second law with
ax = 0.
(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.)
4 answers
Hey, I did the calculations and got
a - 1258.7 N
b - 1.700 N
c - 3350.739 N
But, the website says the answer is wrong.
a - 1258.7 N
b - 1.700 N
c - 3350.739 N
But, the website says the answer is wrong.
wondering how you got so many significant figures out of that. You must believe your calculator
my bad
the fraction from (b) is upside down
will change answers (b) and (c)
heed bob on the sig fig
the fraction from (b) is upside down
will change answers (b) and (c)
heed bob on the sig fig