<\/p>\n\n\n\n
opposing this motion is the static friction force (f_s) between block B and the table. For the system to remain stationary, the tension must be less than or equal to the maximum static friction force. We are interested in the tipping point, so we set them equal.<\/p>\n\n\n\n
T = f_s_max<\/p>\n\n\n\n
The maximum static friction force is calculated as `fHere is the correct answer and explanation for the problem.<\/p>\n\n\n\n
Correct Answer:<\/strong> 3m<\/p>\n\n\n\n To solve this problem, we need to find the minimum mass of block C (m_C) that will prevent block B from sliding. This means the system must be in static equilibrium. In this state, the forces are balanced, and the static friction force is at its maximum value.<\/p>\n\n\n\n First, let’s analyze the forces acting on block A. Block A is being pulled down by gravity and held up by the tension in the rope. For the system to be stationary, the upward tension force (T) must equal the downward gravitational force (weight) of block A.<\/p>\n\n\n\n This tension pulls block B to the right. To prevent block B from sliding, the force of static friction (f_s) acting to the left must be equal to this tension. We are looking for the minimum mass of C, which corresponds to the maximum possible static friction.<\/p>\n\n\n\n The maximum static friction force is calculated as the coefficient_s_max = \u03bc_s * N, where\u03bc_sis the coefficient of static friction andN` is the normal force. The normal force is the upward force exerted by the table, which balances the total weight of the objects on it (blocks B and C).<\/p>\n\n\n\n Total weight on the table: W_BC = (m_B + m_C) * g Now we can set up the main equation by substituting our expressions for T of static friction (\u03bc_s) multiplied by the normal force (N`). The normal force is the upward force exerted by the table on the objects resting on it. In this case, the normal force must support the combined weight of block B and block C.<\/p>\n\n\n\n We can cancel the acceleration due to gravity, m_C`<\/p>\n\n\n\n _s * N = 0.2 * (2m + m_C) * g`<\/p>\n\n\n\n For the system toNow, we substitute the given value for the coefficient of static friction, \u03bc_s = 0.2: To solve for `m_C be in equilibrium, the tension pulling right must equal the maximum static friction pushing left:<\/p>\n\n\n\n T = f_s,max“, we can divide both sides by 0.2:m \/ 0.2 = 2m We can cancel the + m_C5m = 2m + m_C`<\/p>\n\n\n\n Finally, we isolate acceleration due to gravity (g`) from both sides of the equation:<\/p>\n\n\n\n m = 0.2 * (2m +m_C by subtracting 2m from both sides: Now, we solve for m_C:<\/p>\n\n\n\n Therefore, the minimum mass of block C required.2: Therefore, the minimum mass of block C required to keep block B from sliding is 3m<\/strong>.thumb_upthumb_down<\/p>\n\n\n\n Blocks A and B have masses mA and mB, respectively. They are joined by a massless rope that passes over a massless and frictionless pulley. If the coefficient of static friction between block B and the table surface is 0.2, then what is the minimum mass of block C (in terms of the mass m) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-44979","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44979","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=44979"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44979\/revisions"}],"predecessor-version":[{"id":44993,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44979\/revisions\/44993"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44979"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44979"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44979"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation<\/h3>\n\n\n\n
\n
Given m_B = 2m, the normal force is N = (2m + m_C) * g.<\/p>\n\n\n\n\n
T = \u03bc_s * N
m * g = \u03bc_s * (2m + m_C) * g<\/li>\n<\/ul>\n\n\n\n\n
m = 0.2 * (2m + m_C)<\/p>\n\n\n\n
m * g = 0.2 * (2m + m_C) * g<\/p>\n\n\n\n
m_C = 5m – m_C)<\/p>\n\n\n\n\n
m \/ 0.2 = 2m + m_C
to keep block B from sliding is3m.5m = 2m + m_C
2. Subtract 2m from both sides to isolate m_C:
5m – 2m = m_C
m_C = 3m<\/p>\n\n\n\n![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1562.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"