Please answer all 4 questions which are in the red circle. SOLUTIONApply Newton's second law to the sled,with a = 0.F-7.5-F, -oExtract the x-component from this= T + 0 – mg sin 0 = T - (77.0 N)(sin 30.0°) = 0equation to find T. The x-component ofT = 38.5 Nthe normal force is zero, and the sled'sweight is given by mg = 77.0 N.Write the y-component of Newton's> Fy =0 + n – mg cos 0 = n – (77.0 N)(cos 30.0°) = 0second law. The y-component of thetension is zero, so this equation willn = 66.7 Ngive the normal force.LEARN MOREREMARKS Unlike its value on a horizontal surface, n is less than the weight of the sled when the sled ison the slope. This is because only part of the force of gravity (the x-component) is acting to pull the sleddown the slope. The y-component of the force of gravity balances the normal force.QUESTION Consider the same scenario on a hill with a steeper slope.Would the magnitude of the tension in the rope get larger, smaller, or remain the same as before?O The magnitude of the tension force would be greater.O The magnitude of the tension force would be smaller.O The magnitude of the tension force would remain the same.How would the normal force be affected?O The magnitude of the normal force would be greater.O The magnitude of the normal force would be smaller.The magnitude of the normal force would remain the same.BROCTICE IT The magnitude of the normal force would be greater.O The magnitude of the normal force would be smaller.The magnitude of the normal force would remain the same.PRACTICE ITUse the worked example above to help you solve this problem. A sled is tied to a tree on a frictionless,snow-covered hill, as shown in Figure (a). If the sled weighs 67.0 N, find the magnitude of the tensionforceexerted by the rope on the sled and that of the normal force n exerted by the hill on the sled.T =Nn =NEXERCISEHINTS:GETTING STARTEDI'M STUCK!Use the values from PRACTICE IT to help you work this exercise. Suppose a child of weight w climbs ontothe sled. If the tension force is measured to be 60.5 N, find the weight of the child and the magnitude ofthe normal force acting on the sled.W =n%3D

Answered: Image /qna-images/answer/b1a57e16-2701-4f8e-a171-89c2ece957ff.jpg

Use the worked example above to help you solve this problem. A sled is tied to a tree on a frictionless, snow-covered hill, as shown in Figure (a). If the sled weighs 67.0 N, find the magnitude of the tension force T exerted by the rope on the sled and that of the normal force n exerted by the hill on the sled. T = n = EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose a child of weight w climbs onto the sled. If the tension force is measured to be 60.5 N, find the weight of the child and the magnitude of the normal force acting on the sled. = n =

Use the worked example above to help you solve this problem. A sled is tied to a tree on a frictionless, snow-covered hill, as shown in Figure (a). If the sled weighs 67.0 N, find the magnitude of the tension force T exerted by the rope on the sled and that of the normal force n exerted by the hill on the sled. T = n = EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose a child of weight w climbs onto the sled. If the tension force is measured to be 60.5 N, find the weight of the child and the magnitude of the normal force acting on the sled. = n =

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter7: Gravitation

Section: Chapter Questions

Problem 98A

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Concept explainers

Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”

Question

Please answer all 4 questions which are in the red circle.

SOLUTION
Apply Newton's second law to the sled,
with a = 0.
F-7.5-F, -o
Extract the x-component from this
= T + 0 – mg sin 0 = T - (77.0 N)(sin 30.0°) = 0
equation to find T. The x-component of
T = 38.5 N
the normal force is zero, and the sled's
weight is given by mg = 77.0 N.
Write the y-component of Newton's
> Fy =0 + n – mg cos 0 = n – (77.0 N)(cos 30.0°) = 0
second law. The y-component of the
tension is zero, so this equation will
n = 66.7 N
give the normal force.
LEARN MORE
REMARKS Unlike its value on a horizontal surface, n is less than the weight of the sled when the sled is
on the slope. This is because only part of the force of gravity (the x-component) is acting to pull the sled
down the slope. The y-component of the force of gravity balances the normal force.
QUESTION Consider the same scenario on a hill with a steeper slope.
Would the magnitude of the tension in the rope get larger, smaller, or remain the same as before?
O The magnitude of the tension force would be greater.
O The magnitude of the tension force would be smaller.
O The magnitude of the tension force would remain the same.
How would the normal force be affected?
O The magnitude of the normal force would be greater.
O The magnitude of the normal force would be smaller.
The magnitude of the normal force would remain the same.
BROCTICE IT

The magnitude of the normal force would be greater.
O The magnitude of the normal force would be smaller.
The magnitude of the normal force would remain the same.
PRACTICE IT
Use the worked example above to help you solve this problem. A sled is tied to a tree on a frictionless,
snow-covered hill, as shown in Figure (a). If the sled weighs 67.0 N, find the magnitude of the tension
force
exerted by the rope on the sled and that of the normal force n exerted by the hill on the sled.
T =
N
n =
N
EXERCISE
HINTS:
GETTING STARTED
I'M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose a child of weight w climbs onto
the sled. If the tension force is measured to be 60.5 N, find the weight of the child and the magnitude of
the normal force acting on the sled.
W =
n
%3D

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