3.A golf ball is hit of it’s tee, 200m down the fairway.
Draw diagrams to show all the forces acting on the ball
a) When the ball rests on the tree
b)while the club strikes the ball on the tree
c) as the ball is in flight

3A golf ball is hit of its tee 200m down the fairway Draw diagrams to show all the forces acting on the ball a When the ball rests on the tree bwhile the club s class=

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Answers:


3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:


a) Here the ball is under 2 forces:


F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests


-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).



In this case the sum of the forces is 0



b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:


F2, the force of the strike





c) While the ball is in its flight, it is under the following forces:


F1, the force of the lift through the air


-F2, the gravity force

-F3, the force of air resistance, also called drag


F4, the tangencial force of the ball flight




4. Here are the sizes and directions of the resultant forces:


i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:



F=-10N+10N=0 The resulting force is zero



ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:



F=30N-40N=-10N This means the resulting force is 10N applied downwards



iii) In this case the only force applied to the block is -5N applied downwards



iv) Here there are four forces applied to the block.  

In the y-axis we have to forces of the same size but opposite directions:



F1=10N-10N=0 This means the applied force in the y-axis is zero



In the x-axis we have two forces of different size and opposite directions:



F2=-15N+10N=-5N This means the resulting force is applied to the –x-side



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