PHY 140 Practice Quiz #2 Spring 2018 A car is being pulled out of the mud by two
ID: 1576434 • Letter: P
Question
PHY 140 Practice Quiz #2 Spring 2018 A car is being pulled out of the mud by two forces that are applied by the two ropes shown in the drawing. The dashed line in the drawing bisects the 30.0 angle. The magnitude of the force applied by each rope is 2900 newtons (N) (a)How much force would a single rope need to apply to accomplish the same effect as the two forces added together? (b)What angle would the single rope make relative to the dashed line? Problem #2 At a pienic, there is a contest in which hoses are used to shoot water at a beach ball from three directions. As a result, three forces act on the of Fl and F, are F-30.0 newtons and (b) the angle 0 such that the resultant force acting on the ball is zero. ball, P. F. and R3(see drawing). The magnitudes F-70.0 newtons. Determine (a) the magnitude of F3 and 60.0 Problem 3 Which of the following equations are dimensionally correct? Explain your answer b. x1/2 (vt -at)Explanation / Answer
1) Single force should be equal n magnitude to the magnitude of sum of two forces and in the direction of, direction of sum of two forces.
Magnitude of sum of two forces of equal magnitude = 2F cos theta/2, where F is magnotude of each force and Theta is angle between the forces.
single force magnitude = 2*2900*cos 15 = 5602.4 N
Direction of reultant force is along the angle bisector of two equal magnitude forces> Hence single force should be along dashed line.
2) Component of force F2 along east = F2 cos 60 = 35 N
Force F1 is along west and equal to 30 N
componet of f3 along east = F3 cos theta
Net force along east west = 0, hence
35 + F3 cos theta = 30
F3 cos theta = -5 ......1
component of F2 along south = F2 sin 60 = 60.6N
As there is no component of F1 along north-south, component of F3 along north
f3 sin theta = 60.6 N .......2
from 1 and 2
tan theta = -60.6/5
theta = 94.7 deg ( cos theta is -ve and sin theta is +ve, tan theta is between 90 and 180 deg)
form equation 2
F3 sin94.7 = 60.6
F3 = 60.8 N
3) 'b' is dimensionally correct
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