A 52 N force, directed at an angle ? above the horizontal, is applied to a 13.0

Question

A 52 N force, directed at an angle ? above the horizontal, is applied to a 13.0 kg chair sitting on the floor.
(a) For each of the following angles ?, calculate the magnitude of the normal force of the floor on the chair and the horizontal component of the applied force:
(i) 0°
_________N (normal force)
__________N (Fx)

(ii) 30°
___________ N (normal force)
__________ N (Fx)

(iii) 60°
_________N (normal force)
_________N (Fx)
(b) Take the coefficient of static friction between the chair and the floor to be 0.420. Does the chair slide?

For part (i),
No or Yes

For part (ii)?
No or Yes

For part (iii)?
Yes or No

Explanation / Answer

normal force = m*g - F*sin

horizontal component = F*cos

1)normal =127 +0 =127.4 N

horizontal =52 N

2)normal =127.4 - 26 =101.4 N

horizontal =45.033 N

3)normal =82.37 N

horizontal =26 N

for sliding, horizontal force > frictional force
for sliding, horizondal force > frictional force

1) friction force = normal force* =53.5 N

here friction force is greater so no sliding

2)friction force = normal force* =42.6N

here horizondal force is greater so sliding. yes

3)max friction force = normal force* =34.6 N

here horizondal force is less than maximum frictional force so no sliding. no

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