A golf club makes impact with a 0.045 kg golf ball during a drive off the tee. T

Question

A golf club makes impact with a 0.045 kg golf ball during a drive off the tee. The loft angle of the club (the angle between the club face and vertical) is 10 degrees. The normal contact force exerted by the club on the ball is a 7500 N force directed forward and upward 10 degrees above horizontal. The friction force exerted by the club on the ball is 700 N directed forward and downward 80 degrees below horizontal.

a. What is the magnitude of the vector sum of the two club head forces acting on the golf ball?

b. What is the direction of this resultant force?

Explanation / Answer

Here,

mass of golf , m = 0.045 kg

angle = 10 degree

force , F = 7500 N

a) FN = 7500 * (cos(10) i + j * sin(10))

for the friction force

Ff = 700 * (cos(80) i -j * sin(80))

for the vector sum

Fnet = FN + ff

Fnet = 7500 * (cos(10) i + j * sin(10)) + 700 * (cos(80) i -j * sin(80))

Fnet = 7507 i + 613 j N

Fnet = sqrt(7507^2 + 613^2) at arctan(613/7507) degree above horizontal

Fnet = 7532 N at 4.67 degree above horizontal

magnitude of vector sum of the head is 7532 N

b)

the direction of this resulant force is 4.67 degree above horizontal

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